{"id":625664,"date":"2023-09-07T12:03:28","date_gmt":"2023-09-07T12:03:28","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=625664"},"modified":"2023-09-07T12:03:33","modified_gmt":"2023-09-07T12:03:33","slug":"kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/","title":{"rendered":"KC Sinha: Exercise 3.4 – Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923"},"content":{"rendered":"\n\n\n\n\n

Question 2<\/h4>\n\n\n\n

\u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

ax + by = a \u2013 b<\/strong>
bx \u2013 ay = a + b<\/strong>
Sol :
<\/p>\n\n\n\n

a1<\/sub>=a, b1<\/sub>=b, c1<\/sub>=a-b<\/p>\n\n\n\n

a2<\/sub>=b, b2<\/sub>=-a, c2<\/sub>=a+b<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

$\\frac{x}{b(a+b)+a(a-b)}=\\frac{b}{b(a-b)-a(a+b)}-\\frac{-1}{-a^{2}-b^{2}}$<\/p>\n\n\n\n

$\\frac{x}{a b+b^{2}+a^{2}-a b}=\\frac{y}{a b-b^{2}-a^{2}-a b}=\\frac{-1}{-\\left(a^{2}+b^{2}\\right)}$<\/p>\n\n\n\n

$\\frac{x}{a^{2}+b^{2}}=\\frac{y}{-\\left(a^{2}+b^{2}\\right)}=\\frac{1}{a^{2}+b^{2}}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{a^2+b^2}=\\frac{1}{a^2+b^2}&\\frac{y}{-(a^2+b^2)}=\\frac{1}{a^2+b^2}\\\\x=1 & y=-1\\end{array}$&<\/p>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

$\\frac{x}{a}+\\frac{y}{b}=a+b$<\/p>\n\n\n\n

$\\frac{x}{a^{2}}+\\frac{y}{b^{2}}=2, a \\neq 0, b \\neq 0$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{x}{a}+\\frac{y}{b}=a+b$<\/p>\n\n\n\n

$\\frac{b x+a y}{a b}=a+b$<\/p>\n\n\n\n

bx+ay=a2<\/sup>b+ab2<\/sup>…(i)<\/p>\n\n\n\n

$\\frac{x}{a^{2}}+\\frac{y}{b^{2}}=2$<\/p>\n\n\n\n

$\\frac{b^{2} x+a^{2} y}{a^{2} b^{2}}=2$<\/p>\n\n\n\n

b2<\/sup>x+a2<\/sup>y=2a2<\/sup>b2<\/sup>…(ii)<\/p>\n\n\n\n

a1<\/sub>=b, b1<\/sub>=a, c1<\/sub>=a2<\/sup>b+ab2<\/sup><\/p>\n\n\n\n

a2<\/sub>=b2<\/sup>, b2<\/sub>=a2<\/sup>, c2<\/sub>=2a2<\/sup>b2<\/sup><\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}a&\\quad a^2b+ab^2\\\\&\\times \\\\a^2 &\\quad 2a^2b^2\\end{aligned}}=\\frac{y}{\\begin{aligned}a^2b+ab^2 &\\quad b\\\\&\\times \\\\2a^2b^2&\\quad b^2\\end{aligned}}=\\frac{-1}{\\begin{aligned}b&\\quad a\\\\&\\times\\\\b^2&\\quad a^2\\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{2 a^{3} b^{2}-a^{4} b-a^{3} b^{2}}=\\frac{y}{a^{2} b^{3}+a b^{4}-2 a^{2} b^{3}}=\\frac{-1}{a^{2} b-a b^{2}}$<\/p>\n\n\n\n

$\\frac{x}{a^{3} b^{2}-a^{4} b}=\\frac{y}{a b^{4}-a^{2} b^{3}}=\\frac{-1}{a^{2} b-a b^{2}}$<\/p>\n\n\n\n

$\\frac{x}{a^{3} b(b-a)}=\\frac{y}{a b^{3}(b-a)}=\\frac{-1}{-\\left(a b^{2}-a^{2} b\\right)}$<\/p>\n\n\n\n

$\\frac{x}{a^{3} b(b-a)}=\\frac{y}{a b^{3}(b-4)}=\\frac{1}{a b(b-a)}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{a^{3} b(b-a)}=\\frac{1}{a b(b-a)}&\\frac{y}{a b^{3}(b-a)}-\\frac{1}{a b(b-a)}\\\\x=\\frac{a^3b(b-a)}{ab(b-a)}&y=\\frac{ab^3(b-a)}{ab(b-a)}\\end{array}$<\/p>\n\n\n\n

$x=a^{2},y=b^{2}$<\/p>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

x+y=a+b<\/strong><\/p>\n\n\n\n

az+by=a2<\/sup>-b2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a1<\/sub>=1, b1<\/sub>=-1, c1<\/sub>=a+b<\/p>\n\n\n\n

a2<\/sub>=a, b2<\/sub>=b, c2<\/sub>=a2<\/sup>-b2<\/sup><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

$\\frac{x}{-a^{2}+b^{2}-a b-b^{2}}=\\frac{y}{a^{2}+a b-a^{2}+b^{2}}=\\frac{-1}{b+a}$<\/p>\n\n\n\n

$\\frac{x}{-a(a+b)}=\\frac{y}{b(a+b)}=\\frac{-1}{a+b}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{-a(a+b)}=\\frac{-1}{a+b}&\\frac{y}{b(a+b)}=\\frac{-1}{a+b}\\\\x=a& y=-b\\end{array}$<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{2 x}{a}+\\frac{y}{b}=2$<\/p>\n\n\n\n

$\\frac{x}{a}-\\frac{y}{b}=4, a \\neq 0, b \\neq 0$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{2 x}{a}+\\frac{y}{b}=2$<\/p>\n\n\n\n

$\\frac{2 b x+ay}{a b}=2$<\/p>\n\n\n\n

2bx+ay=2ab…(i)<\/p>\n\n\n\n

$\\frac{x}{a}-\\frac{y}{b}=4$<\/p>\n\n\n\n

$\\frac{b x-a y}{a b}=4$<\/p>\n\n\n\n

bx-ay=4ab..(ii)<\/p>\n\n\n\n

a1<\/sub>=2b, b1<\/sub>=a, c1<\/sub>=2ab<\/p>\n\n\n\n

a2<\/sub>=b, b2<\/sub>=-a, c2<\/sub>=4ab<\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}a\\quad&2ab\\\\\\times & \\\\-a\\quad&4ab\\end{aligned}}=\\frac{y}{\\begin{aligned}2ab&\\quad 2b\\\\ &\\times \\\\4ab& \\quad b\\end{aligned}}=\\frac{-1}{\\begin{aligned}2b& \\quad a\\\\ &\\times \\\\b&\\quad -a \\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{4 a^{2} b-\\left(-2 a^{2} b\\right)}=\\frac{y}{2 a b^{2}-8 a b^{2}}=\\frac{-1}{-2 a b-a b}$<\/p>\n\n\n\n

$\\frac{x}{6 a^{2} b}=\\frac{y}{-6 a b^{2}}=\\frac{-1}{-3 a b}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{6 a^{2} b}=\\frac{1}{3 a b}&\\frac{y}{-6 a b^{2}}=\\frac{1}{3 a 1}\\\\x=\\frac{6 a^{2} b}{3ab}&y=\\frac{-6ab}{3ab}\\\\x=2a& y=-2b\\end{array}$<\/p>\n\n\n\n

(v) <\/strong><\/p>\n\n\n\n

2ax+3by=a+2b<\/strong><\/p>\n\n\n\n

3ax+2by=2a+b<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a1<\/sub>=2a, b1<\/sub>=3b, c1<\/sub>=a+2b<\/p>\n\n\n\n

a2<\/sub>=3a, b2<\/sub>=2b, c2<\/sub>=2a+b<\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}3b&\\quad a+2b\\\\&\\times \\\\2b&\\quad 2a+b\\end{aligned}}=\\frac{y}{\\begin{aligned}a+2b&\\quad 2a\\\\&\\times \\\\2a+b& \\quad 3a\\end{aligned}}=\\frac{-1}{\\begin{aligned}2a&\\quad 3b\\\\&\\times \\\\3a&\\quad 2b \\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{3 b(2 a+b)-2 b(a+2 b)}=\\frac{y}{3 a(a+2 b)-2 a(2 a+b)}=\\frac{-1}{4 a b-9 a b}$<\/p>\n\n\n\n

$\\frac{x}{4 a b-b^{2}}=\\frac{y}{4 a b-a^{2}}=\\frac{1}{5 a b}$<\/p>\n\n\n\n

$\\frac{x}{b(4 a-b)}=\\frac{y}{a(4 b-a)}=\\frac{1}{5 a b}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{b(4a-b)}&\\frac{1}{5ab}\\\\\\frac{y}{a(4b-a)}&\\frac{1}{5ab}\\\\x=\\frac{4a-b}{5a}&y=\\frac{4b-a}{5b}\\end{array}$<\/p>\n\n\n\n

(vi)<\/strong><\/p>\n\n\n\n

$\\frac{x}{a}+\\frac{y}{b}=2$<\/strong>
ax + by = a2<\/sup>\u2013b2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{x}{a}+\\frac{y}{b}=2$<\/p>\n\n\n\n

$\\frac{b x+a y}{a b}=2$<\/p>\n\n\n\n

bx+ay=2ab<\/p>\n\n\n\n

a1<\/sub>=b, b1<\/sub>=a, c1<\/sub>=2ab<\/p>\n\n\n\n

a2<\/sub>=a, b2<\/sub>=-b, c2<\/sub>=a2<\/sup>-b2<\/sup><\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}a&\\quad 2ab\\\\&\\times \\\\-b&\\quad a^2-b^2\\end{aligned}}=\\frac{y}{\\begin{aligned}2ab&\\quad b\\\\& \\times \\\\a^2-b^2& \\quad a\\end{aligned}}=\\frac{-1}{\\begin{aligned}b& \\quad a\\\\ &\\times \\\\a&\\quad -b\\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{a\\left(a^{2}-b^{2}\\right)+2 a b^{2}}=\\frac{y}{2 a^{2} b-b\\left(a^{2}-b^{2}\\right)}=\\frac{-1}{-b^{2}-a^{2}}$<\/p>\n\n\n\n

$\\frac{x}{a^{3}-a b^{2}+2 a b^{2}}=\\frac{y}{2 a^{2} b-a^{2} b+b^{3}}=\\frac{-1}{-\\left(b^{2}+a^{2}\\right)}$<\/p>\n\n\n\n

$\\frac{x}{a^{3}+a b^{2}}=\\frac{y}{a^{2} b+b^{3}}=\\frac{1}{a^{2}+b^{2}}$<\/p>\n\n\n\n

$\\frac{x}{a\\left(a^{2}+b^{2}\\right)}=\\frac{y}{b\\left(a^{2}+b^{2}\\right)}=\\frac{1}{a^{2}+b^{2}}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{a\\left(a^{2}+b^{2}\\right)}=\\frac{1}{a^{2}+b^{2}}&\\frac{y}{b\\left(a^{2}+b^{x}\\right)}=\\frac{1}{a^{2}+b^{2}}\\\\x=a& y=b\\end{array}$<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u092f \u0915\u094b \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0939\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

a(x + y) + b(x \u2013 y) =a2<\/sup>\u2013 ab + b2<\/sup><\/strong>
a(x + y) \u2013 b(x \u2013 y) = a2<\/sup> + ab + b2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947,<\/p>\n\n\n\n

a(x+y)+b(x-y)=a2<\/sup>\u2013ab+b2<\/sup><\/p>\n\n\n\n

ax+ay+bx-by=a2<\/sup>-ab+b2<\/sup><\/p>\n\n\n\n

(a+b)x+(a-b)y=a2<\/sup>-ab+b2<\/sup><\/p>\n\n\n\n

\u0907\u0938\u0940\u092a\u094d\u0930\u0915\u093e\u0930 \u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u0938\u0947<\/p>\n\n\n\n

a(x+y)\u2013b(x\u2013y)=a2<\/sup>+ab+b2<\/sup><\/p>\n\n\n\n

ax+ay-bx+by=a2<\/sup>+ab+b2<\/sup><\/p>\n\n\n\n

(a-b)x+(a+b)y=a2<\/sup>+ab+b2<\/sup><\/p>\n\n\n\n

a1<\/sub>=a+b, b1<\/sub>=a-b, c1<\/sub>=a2<\/sup>-ab+b2<\/sup><\/p>\n\n\n\n

a2<\/sub>=a-b, b2<\/sub>=a+b, c2<\/sub>=a2<\/sup>+ab+b2<\/sup><\/p>\n\n\n\n

$\\dfrac{x}{\\begin{aligned}a-b& \\quad a^2-ab+b^2 \\\\ & \\times \\\\a+b & \\quad a^2+ab+b^2 \\end{aligned}}=\\dfrac{y}{\\begin{aligned}a^2-ab+b^2 & \\quad a+b\\\\ & \\times \\\\ a^2+ab+b^2 & \\quad a-b \\end{aligned}}=\\dfrac{-1}{\\begin{aligned}a+b & \\quad a-b \\\\ & \\times \\\\ a-b & \\quad a+b\\end{aligned}}$<\/p>\n\n\n\n

$\\dfrac{x}{(a-b)(a^2+ab+b^2)-(a+b)(a^2-ab+b^2)}=\\dfrac{y}{(a-b)(a^2-ab+b^2)-(a+b)(a^2+ab+b^2)}=\\dfrac{-1}{(a+b)^2-(a-b)^2}$<\/p>\n\n\n\n

$\\dfrac{x}{\\left(a^{3}-b^{3}\\right)-\\left(a^{3}+b^{3}\\right)}=\\dfrac{y}{a^3-a^2b+ab^2-a^2b+ab^2-b^3-a^3-a^2b-ab^2-a^2b-ab^2-b^3}=\\dfrac{-1}{4 a b}$<\/p>\n\n\n\n

$\\frac{x}{a^{3}-b^{3}-a^{3}-b^{3}}=\\frac{y}{-4 a^{2} b-2 b^{3}}=\\frac{-1}{4 a b}$<\/p>\n\n\n\n

$\\frac{x}{-2 b^{3}}=\\frac{y}{-2 b\\left(2 a^{2}+b^{2}\\right)}=\\frac{-1}{4 a b}$<\/p>\n\n\n\n

$\\frac{x}{-2 b^{3}}=\\frac{-1}{4 a b}$<\/p>\n\n\n\n

$x=\\frac{2b^3}{2^{4ab}}=\\frac{b^2}{2a}$<\/p>\n\n\n\n

$\\frac{y}{-2 b\\left(2a^{2}+b^{2}\\right)}=\\frac{-1}{4 a b}$<\/p>\n\n\n\n

$y=\\frac{2 b\\left(2 a^{2}+b^{2}\\right)}{4a b}$<\/p>\n\n\n\n

$y=\\frac{2 a^{2}+b^{2}}{2 a}$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

\u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0938\u0947 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915-\u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e \u092e\u0947 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0939\u0948, \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0928\u0939\u0940 \u0939\u0948 \u0924\u0925\u093e \u0915\u093f\u0938\u0924\u093e \u0905\u092a\u0930\u093f\u092e\u093f\u0924\u093f \u0930\u0942\u0902\u092a \u0938\u0947 \u0905\u0928\u0947\u0915 \u0939\u0932 \u0939\u0948?<\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

x-3y-7=0<\/p>\n\n\n\n

3x-3y-15=0<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a1<\/sub>=1, b1<\/sub>=-3, c1<\/sub>=-7<\/p>\n\n\n\n

a2<\/sub>=3, b2<\/sub>=-3, c2<\/sub>=-15<\/p>\n\n\n\n

(ii) <\/strong><\/p>\n\n\n\n

2x+y=5<\/p>\n\n\n\n

3x+2y=8<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

3x-5y=20<\/p>\n\n\n\n

6x-10y=40<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a1<\/sub>=3, b1<\/sub>=-5, c1<\/sub>=20<\/p>\n\n\n\n

a2<\/sub>=6, b2<\/sub>=-10, c2<\/sub>=40<\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}-5 & \\quad 20\\\\ & \\times \\\\ -10 & \\quad 40\\end{aligned}}=\\frac{y}{\\begin{aligned}20 & \\quad 3 \\\\ & \\times \\\\40 & \\quad 6 \\end{aligned}}=\\frac{-1}{\\begin{aligned}3 &\\quad -5\\\\ & \\times \\\\ 6 & \\quad -10\\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{-200+200}=\\frac{y}{120-120}=\\frac{-1}{-30+30}$<\/p>\n\n\n\n

$\\frac{x}{0}=\\frac{y}{0}=-\\frac{1}{0}$ \u0905\u0928\u0947\u0915 \u0939\u0932<\/p>\n\n\n\n

(iv) <\/strong><\/p>\n\n\n\n

x-3y-3=0<\/strong><\/p>\n\n\n\n

3x-9y-2=0<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a1<\/sub>=1, b1<\/sub>=-3, c1<\/sub>=-3<\/p>\n\n\n\n

a2<\/sub>=3, b2<\/sub>=-9, c2<\/sub>=-2<\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}-3 & \\quad -3\\\\ & \\times \\\\-9 & \\quad -2\\end{aligned}}=\\frac{y}{\\begin{aligned}-3& \\quad 1 \\\\ & \\times \\\\ -2 & \\quad 3\\end{aligned}}=\\frac{1}{\\begin{aligned}1 & \\quad -3 \\\\ & \\times \\\\ 3 & \\quad -9\\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{6-18}=\\frac{y}{-9+2}=\\frac{1}{-9+9}$<\/p>\n\n\n\n

$\\frac{x}{-12}=\\frac{y}{-7}=\\frac{1}{0}$ \u0915\u094b\u0908 \u0939\u0932 \u0928\u0939\u0940 <\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u092f \u0915\u094b \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0939\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

$\\frac{5}{x+y}-\\frac{2}{x-y}+1=0$<\/p>\n\n\n\n

$\\frac{15}{x+y}+\\frac{7}{x-y}-10=0$<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e $\\frac{1}{x+y}=u, \\frac{1}{x-y}=v$<\/p>\n\n\n\n

5u-2v+1=0<\/p>\n\n\n\n

15u+7v-10=0<\/p>\n\n\n\n

$\\frac{u}{\\begin{aligned}-2&\\quad 1\\\\ & \\times \\\\7& \\quad -10\\end{aligned}}=\\frac{v}{\\begin{aligned}1 &\\quad 5\\\\ & \\times \\\\-10 & \\quad 15\\end{aligned}}=\\frac{1}{\\begin{aligned}5 & \\quad -2\\\\ & \\times \\\\15& \\quad 7\\end{aligned}}$<\/p>\n\n\n\n

$\\frac{u}{20-7}=\\frac{v}{15+50}=\\frac{1}{35+30}$<\/p>\n\n\n\n

$\\frac{u}{13}=\\frac{v}{65}=\\frac{1}{65}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{u}{13}=\\frac{1}{65}&\\frac{v}{65}=\\frac{1}{65}\\\\u=\\frac{1}{5}& v=1\\end{array}$<\/p>\n\n\n\n

$\\because \\frac{1}{x+y}=u , \\frac{1}{x-y}=v$<\/p>\n\n\n\n

x+y=5, x-y=1<\/p>\n\n\n\n

a1<\/sub>=1, b1<\/sub>=1, c1<\/sub>=5<\/p>\n\n\n\n

a2<\/sub>=1, b2<\/sub>=-1, c2<\/sub>=1<\/p>\n\n\n\n

$\\frac{x}{\\begin{aligned}1&\\quad 5\\\\ & \\times \\\\-1 & \\quad 1\\end{aligned}}=\\frac{y}{\\begin{aligned}5 & \\quad 1\\\\ & \\times \\\\ 1 & \\quad 1\\end{aligned}}=\\frac{-1}{\\begin{aligned}1& \\quad 1\\\\ & \\times \\\\1 & \\quad -1 \\end{aligned}}$<\/p>\n\n\n\n

$\\frac{x}{1+5}=\\frac{y}{5-1}=\\frac{-1}{-1-1}$<\/p>\n\n\n\n

$\\frac{x}{6}=\\frac{y}{4}=\\frac{-1}{-2}$<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x}{6}=\\frac{1}{2}& \\frac{y}{4}=\\frac{1}{2}\\\\ x=\\frac{6}{3}=3& y=\\frac{4}{5}=2\\end{array}$<\/p>\n\n\n\n

(ii) ax-ay=2<\/strong><\/p>\n\n\n\n

(a-1)x+(a+1)y=2(a2<\/sup>+1)<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

\u092f\u0926\u093f 2 \u092a\u0947\u0902\u0938\u093f\u0932 \u0914\u0930 3 \u0930\u092c\u0921\u093c \u0915\u0940 \u0915\u0940\u092e\u0924 9 \u0930\u0942 \u0939\u094b \u0914\u0930 4 \u092a\u0949\u0938\u093f\u0932 \u0914\u0930 6 \u0930\u092c\u0921\u093c \u0915\u0940 \u0915\u0940\u092e\u0924 18 \u0930\u0942. \u0939\u094b\u0902 \u0924\u094b \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092a\u0947\u0902\u0938\u093f\u0932 \u0914\u0930 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0930\u092c\u0921\u093c \u0915\u0940 \u0915\u0940\u092e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e 1 \u092a\u0947\u0902\u0938\u093f\u0932 \u0915\u0940 \u0915\u0940\u092e\u0924=x<\/p>\n\n\n\n

1 \u0930\u092c\u0921\u093c \u0915\u0940 \u0915\u0940\u092e\u0924=y<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

2x+3y=9..(i)<\/p>\n\n\n\n

4x+6y=18..(ii)<\/p>\n\n\n\n

$\\frac{a_1}{a_{2}}=\\frac{2}{4}=\\frac{1}{2}$, $\\frac{b_{1}}{b_{2}}=\\frac{3}{6}$, $\\frac{c_{1}}{c_{2}}=\\frac{9}{18}=\\frac{1}{2}$<\/p>\n\n\n\n

$\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$ \u0905\u0928\u0947\u0915 \u0939\u0932<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

\u0926\u094b \u0930\u0947\u0932\u0917\u093e\u0921\u093c\u093f\u092f\u094b\u0902 \u0926\u094d\u0935\u093e\u0930\u093e \u0924\u092f \u092e\u093e\u0930\u094d\u0917 \u0915\u094b \u0938\u092e\u0940\u0915\u0930\u0923\u094b\u0902 x+2y-4=0 \u0914\u0930 2 x+4y-12=0 \u0938\u0947 \u092a\u094d\u0930\u0915\u091f \u0915\u093f\u092f\u093e \u0917\u092f\u093e \u0939\u0948 \u0964 \u0915\u094d\u092f\u093e \u092f\u0947 \u092e\u093e\u0930\u094d\u0917 \u090f\u0915-\u0926\u0941\u0938\u0930\u0947 \u0915\u094b \u0915\u093e\u091f\u0947\u0902\u0917\u0947 ?<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

x+2y-4=0<\/p>\n\n\n\n

2x+4y-12=0<\/p>\n\n\n\n

a1<\/sub>=1, b1<\/sub>=2, c1<\/sub>=-4<\/p>\n\n\n\n

a2<\/sub>=2, b2<\/sub>=4, c2<\/sub>=-12<\/p>\n\n\n\n

\\frac{a_{1}}{a_{2}}=\\frac{1}{2}, $\\frac{b_{1}}{b_{2}}=\\frac{2}{4}$, $\\frac{c_{1}}{c_{2}}=\\frac{-{4}}{-12}$<\/p>\n\n\n\n

$\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{3}}$ \u0915\u094b\u0908 \u0939\u0932 \u0928\u0939\u0940 <\/p>\n\n\n\n

\u0905\u0924\u0903 \u0926\u094b\u0928\u094b \u092e\u093e\u0930\u094d\u0917 \u090f\u0915-\u0926\u0942\u0938\u0930\u0947 \u0915\u094b \u0928\u0939\u0940 \u0915\u093e\u091f\u0947\u0902\u0917\u0947\u0964<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

\u0926\u094b \u0906\u0926\u092e\u093f\u092f\u094b \u0915\u0940 \u0906\u092e\u0926\u0928\u0940 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 9: 7 \u0939\u0948 \u0914\u0930 \u0909\u0928\u0915\u0947 \u0916\u0930\u094d\u091a \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 4: 3 \u0939\u0948\u0964 \u092f\u0926\u093f \u0909\u0928\u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092a\u094d\u0930\u0924\u093f \u092e\u093e\u0939 2000 \u0930\u0941 \u092c\u091a\u093e\u0924\u093e \u0939\u0948 \u0924\u094b \u0909\u0928\u0915\u0940 \u092e\u093e\u0938\u093f\u0915 \u0906\u092f \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0926\u094b\u0928\u094b \u0935\u094d\u092f\u0915\u094d\u0924\u093f\u092f\u094b \u0915\u0940 \u092e\u093e\u0938\u093f\u0915 \u0906\u092f \u0915\u094d\u0930\u092e\u0936\u0903 9x \u0924\u0925\u093e 7x \u0939\u0948\u0964<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0935\u094d\u092f\u0915\u094d\u0924\u093f\u092f\u094b \u0915\u093e \u092e\u093e\u0938\u093f\u0915 \u0916\u0930\u094d\u091a \u0915\u094d\u0930\u092e\u0936\u0903 4y \u0924\u0925\u093e 3y \u0939\u0948\u0964<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947 ,<\/p>\n\n\n\n

9x-4y=2000..(i)\u00d73<\/p>\n\n\n\n

7x-3y=2000..(ii)\u00d74<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}27x-12y&=6000\\\\28x-12y&=8000\\\\-\\phantom{28x}+\\phantom{12y}&\\phantom{=}-\\phantom{8000}\\\\ \\hline -x=&-2000\\end{aligned}$<\/p>\n\n\n\n

x=2000<\/p>\n\n\n\n

\u092a\u0939\u0932\u0947 \u0935\u094d\u092f\u0915\u094d\u0924\u093f \u0915\u0940 \u092e\u093e\u0938\u093f\u0915 \u0906\u092f=9x<\/p>\n\n\n\n

=9\u00d72000<\/p>\n\n\n\n

=18000<\/p>\n\n\n\n

\u0926\u0942\u0938\u0930\u0947 \u0935\u094d\u092f\u0915\u094d\u0924\u093f \u0915\u0940 \u092e\u093e\u0938\u093f\u0915 \u0906\u092f=7x<\/p>\n\n\n\n

=7\u00d72000<\/p>\n\n\n\n

=14000<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u0926\u094b \u0905\u0902\u0915\u094b\u0902 \u0915\u0940 \u090f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e \u090f\u0935\u0902 \u0905\u0902\u0915\u094b\u0902 \u0915\u0947 \u092a\u0932\u091f\u0928\u0947 \u092a\u0930 \u092c\u0928\u0940 \u0938\u0902\u0916\u094d\u092f\u093e \u0915\u093e \u092f\u094b\u0917 66 \u0939\u0948\u0964 \u092f\u0926\u093f \u0938\u0902\u0916\u094d\u092f\u093e \u0915\u0947 \u0905\u0902\u0915\u094b\u0902 \u0915\u093e \u0905\u0902\u0924\u0930 2 \u0939\u094b \u0924\u094b \u0938\u0902\u0916\u094d\u092f\u093e \u092c\u0924\u093e\u0907\u092f\u0947 \u0964 \u0910\u0938\u0940 \u0915\u093f\u0924\u0928\u0940 \u0938\u0902\u0916\u094d\u092f\u093e\u092f\u0947\u0902 \u0939\u094b\u0902\u0917\u0940 ?<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0907\u0915\u093e\u0908 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=x<\/p>\n\n\n\n

\u0926\u0939\u093e\u0908 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=y<\/p>\n\n\n\n

\u0926\u094b \u0905\u0902\u0915\u094b \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=10y+x<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

(10y+x)+(10x+y)=66<\/p>\n\n\n\n

11x+11y=66<\/p>\n\n\n\n

11(x+y)=66<\/p>\n\n\n\n

11(x+y)=66<\/p>\n\n\n\n

$x+y=\\frac{66}{11}$<\/p>\n\n\n\n

x+y=6..(i)<\/p>\n\n\n\n

x-y=2  \u092f\u093e y-x=2 <\/p>\n\n\n\n

Case-I<\/p>\n\n\n\n

$\\begin{aligned}x+y&=6..(i)\\\\x-y&=2..(ii)\\\\2x&=8\\end{aligned}$<\/p>\n\n\n\n

x=4<\/p>\n\n\n\n

x \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

x+y=6<\/p>\n\n\n\n

4+y=6<\/p>\n\n\n\n

y=2<\/p>\n\n\n\n

\u0926\u094b \u0905\u0902\u0915\u094b \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=10y+x<\/p>\n\n\n\n

=10(2)+4<\/p>\n\n\n\n

=20+4<\/p>\n\n\n\n

=24<\/p>\n\n\n\n

CASE-II<\/p>\n\n\n\n

$\\begin{aligned}x+y&=6..(i)\\\\y-x&=2..(ii)\\\\2y&=8\\end{aligned}$<\/p>\n\n\n\n

y=4<\/p>\n\n\n\n

y \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

x+y=6<\/p>\n\n\n\n

x+4=6<\/p>\n\n\n\n

x=2<\/p>\n\n\n\n

\u2234 \u0926\u094b \u0905\u0902\u0915\u094b \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=10\u00d74+2<\/p>\n\n\n\n

=40+2<\/p>\n\n\n\n

=42<\/p>\n\n\n\n

42 \u0914\u0930 24 \u0926\u094b \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 \u0939\u0948\u0964<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 \u092d\u093f\u0928\u094d\u0928 \u0915\u0947 \u0905\u0902\u0936 \u092e\u0947\u0902 1 \u091c\u094b\u0921\u093c\u093e \u091c\u093e\u092f \u0914\u0930 \u0939\u0930 \u0938\u0947 1 \u0918\u091f\u093e\u092f\u093e \u091c\u093e\u092f \u0924\u094b \u0935\u0939 1 \u0939\u094b \u091c\u093e\u0924\u093e \u0939\u0948\u0964 \u092f\u0926\u093f \u0939\u0930 \u092e\u0947\u0902 1 \u091c\u094b\u0921\u093c\u0924\u0947 \u0939\u0948, \u0924\u094b \u092f\u0939 $\\frac{1}{2}$ \u0939\u094b \u091c\u093e\u0924\u093e \u0939\u0948\u0964 \u092d\u093f\u0928\u094d\u0928 \u0915\u094d\u092f\u093e \u0939\u0948 ?<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u092d\u093f\u0928\u094d\u0928 \u0915\u093e \u0905\u0902\u0936=x<\/p>\n\n\n\n

\u092d\u093f\u0928\u094d\u0928 \u0915\u093e \u0939\u0930=y<\/p>\n\n\n\n

\u092d\u093f\u0928\u094d\u0928$=\\frac{x}{y}$<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x+1}{y-1}=1&\\frac{x}{y+1}=\\frac{1}{2}\\\\x+1=y-1&2x=y+1\\\\x-y=-2..(i)&2x-y=1..(ii)\\end{array}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}x-y&=-2\\\\2x-y&=1\\\\-\\phantom{2x}+\\phantom{y}&\\phantom{=}-\\phantom{1}\\\\ \\hline -x=-3\\end{aligned}$<\/p>\n\n\n\n

x=3<\/p>\n\n\n\n

x \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930<\/p>\n\n\n\n

x-y=-2<\/p>\n\n\n\n

3-y=-2<\/p>\n\n\n\n

-y=-2-3<\/p>\n\n\n\n

y=5<\/p>\n\n\n\n

\u2234\u092d\u093f\u0928\u094d\u0928 $=\\frac{x}{y}=\\frac{3}{5}$<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

5 \u0928\u093e\u0930\u0902\u0917\u0940 \u0914\u0930 3 \u0938\u0947\u092c \u0915\u0940 \u0915\u094d\u0930\u0947\u092e\u0924 35 \u0930\u0941 \u0924\u0925\u093e 2 \u0928\u093e\u0930\u0902\u0917\u0940 \u0914\u0930 4 \u0938\u0947\u092c \u0915\u0940 \u0915\u0940\u092e\u0924 28 \u0930\u0941 \u0939\u0948 \u0964 \u090f\u0915 \u0928\u093e\u0902\u0930\u0917\u0940 \u0914\u0930 \u090f\u0915 \u0938\u0947\u092c \u0915\u0940 \u0915\u0940\u092e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e 1 \u0928\u093e\u0930\u0902\u0917\u0940 \u0915\u0940 \u0915\u0940\u092e\u0924=x<\/p>\n\n\n\n

1 \u0938\u0947\u092c \u0915\u0940 \u0915\u0940\u092e\u0924=y<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

5x+3y=35..(i)\u00d72<\/p>\n\n\n\n

2x+4y=28<\/p>\n\n\n\n

2(x+2y)=28<\/p>\n\n\n\n

x+2y=14…(ii)\u00d72<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}10x+6y&=70\\\\3x+6y&=42\\\\ -\\phantom{3x}-\\phantom{6y}&\\phantom{=}-\\phantom{42}\\\\ \\hline {7x=28}\\end{aligned}$<\/p>\n\n\n\n

x=4<\/p>\n\n\n\n

x \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

x+2y=14<\/p>\n\n\n\n

4+2y=14<\/p>\n\n\n\n

2y=10<\/p>\n\n\n\n

$y=\\frac{10}{2}$<\/p>\n\n\n\n

y=5<\/p>\n\n\n\n

\u2234 \u090f\u0915 \u0928\u093e\u0930\u0902\u0917\u0940 \u0915\u0940 \u0915\u0940\u092e\u0924=4<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u0915\u093f\u0938\u0940 \u091b\u093e\u0924\u094d\u0930\u093e\u0935\u093e\u0938 \u092e\u0947 \u092e\u093e\u0938\u093f\u0915 \u0935\u094d\u092f\u092f \u0915\u093e \u090f\u0915 \u092d\u093e\u0917 \u0928\u093f\u092f\u0924 \u0939\u0948 \u0924\u0925\u093e \u0936\u0947\u0937 \u0907\u0938 \u092a\u0930 \u0928\u093f\u0930\u094d\u092d\u0930 \u0939\u0948 \u0915\u093f \u091b\u093e\u0924\u094d\u0930 \u0928\u0947 \u0915\u093f\u0924\u0928\u0947 \u0926\u093f\u0928 \u092d\u094b\u091c\u0928 \u0915\u093f\u092f\u093e \u0939\u0948? \u091c\u092c \u090f\u0915 \u091b\u093e\u0924\u094d\u0930 A \u0915\u094b 20 \u0926\u093f\u0928 \u092d\u094b\u091c\u0928 \u0915\u0930\u0924\u093e \u0939\u0948, 1000 \u0930\u0941 \u0926\u0947\u0928\u093e \u092a\u0921\u093c\u0924\u093e \u0939\u0948 \u091c\u092c\u0915\u093f B \u0915\u094b \u091c\u094b 26 \u0926\u093f\u0928 \u092d\u094b\u091c\u0928 \u0915\u0930\u0924\u093e \u0939\u0948 1180 \u0930\u0941 \u0926\u0947\u0928\u093e \u092a\u0921\u093c\u0924\u093e \u0939\u0948 \u0964 \u0928\u093f\u092f\u0924 \u092e\u093e\u0938\u093f\u0915 \u0935\u094d\u092f\u092f \u0924\u0925\u093e \u090f\u0915 \u0926\u093f\u0928 \u0915\u0947 \u092d\u094b\u091c\u0928 \u0915\u093e \u092e\u0942\u0932\u094d\u092f \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u091b\u093e\u0924\u094d\u0930\u093e\u0935\u093e\u0938 \u0928\u093f\u092f\u0924 \u092e\u093e\u0938\u093f\u0915 \u0935\u094d\u092f\u092f=x<\/p>\n\n\n\n

\u0924\u0925\u093e \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0926\u093f\u0928 \u0915\u0947 \u092d\u094b\u091c\u0928 \u0915\u093e \u092e\u0942\u0932\u094d\u092f=y<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}x+20 y&=1000..(i)\\\\x+26y&=1180..(ii)\\\\ -\\phantom{x}-\\phantom{26y}&\\phantom{=}-\\\\ \\hline-6y&=-180\\end{aligned}$<\/p>\n\n\n\n

$y=\\frac{180}{6}=30$<\/p>\n\n\n\n

y \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

x+20y=1000<\/p>\n\n\n\n

x+20(30)=1000<\/p>\n\n\n\n

x+600=1000<\/p>\n\n\n\n

x=400<\/p>\n\n\n\n

\u2234\u091b\u093e\u0924\u094d\u0930\u093e\u0935\u093e\u0938 \u0915\u093e \u0928\u093f\u092f\u0924 \u092e\u093e\u0938\u093f\u0915 \u0935\u094d\u092f\u092f=400<\/p>\n\n\n\n

\u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0926\u093f\u0928 \u0915\u0947 \u092d\u094b\u091c\u0928 \u0915\u093e \u092e\u0942\u0932\u094d\u092f=30<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

\u0915\u093f\u0938\u0940 \u092d\u093f\u0928\u094d\u0928 \u0915\u0947 \u0905\u0902\u0936 \u0938\u0947 1 \u0918\u091f\u0928\u0947 \u092a\u0930 \u0935\u0939 $\\frac{1}{3}$ \u0939\u094b \u091c\u093e\u0924\u093e \u0939\u0948 \u0914\u0930 \u0939\u0930 \u092e\u0947\u0902 8 \u091c\u094b\u0921\u093c\u0928\u0947 \u092a\u0930 $\\frac{1}{4}$ \u0939\u094b \u091c\u093e\u0924\u093e \u0939\u0948\u0964 \u092d\u093f\u0928\u094d\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u092d\u093f\u0928\u094d\u0928 \u0915\u093e \u0905\u0902\u0936=x<\/p>\n\n\n\n

\u092d\u093f\u0928\u094d\u0928 \u0915\u093e \u0939\u0930=y<\/p>\n\n\n\n

\u092d\u093f\u0928\u094d\u0928$=\\frac{x}{y}$<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947,<\/p>\n\n\n\n

$\\begin{array}{l|l}\\frac{x-1}{4}=\\frac{1}{3}&\\frac{x}{y+8}=\\frac{1}{4}\\\\3 x-3=y&4 x=y+8\\\\3 x-y=3..(i)&4 x-y=8..(ii)\\end{array}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947<\/p>\n\n\n\n

$\\begin{aligned}3 x-y&=3 \\\\4 x-y&=8\\\\-\\phantom{4 x}+\\phantom{y}&\\phantom{=}-\\phantom{8} \\\\ \\hline -x&=-5 \\end{aligned}$<\/p>\n\n\n\n

x=5<\/p>\n\n\n\n

x \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

3x-y=3<\/p>\n\n\n\n

3(5)-y=3<\/p>\n\n\n\n

15-y=3<\/p>\n\n\n\n

-y=3-15<\/p>\n\n\n\n

y=12<\/p>\n\n\n\n

\u2234\u092d\u093f\u0928\u094d\u0928$=\\frac{x}{y}=\\frac{5}{12}$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 2 \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947: ax + by = a \u2013 bbx \u2013 ay = a + bSol : a1=a, b1=b, c1=a-b a2=b, b2=-a, c2=a+b $\\frac{x}{b(a+b)+a(a-b)}=\\frac{b}{b(a-b)-a(a+b)}-\\frac{-1}{-a^{2}-b^{2}}$ $\\frac{x}{a b+b^{2}+a^{2}-a b}=\\frac{y}{a b-b^{2}-a^{2}-a b}=\\frac{-1}{-\\left(a^{2}+b^{2}\\right)}$ $\\frac{x}{a^{2}+b^{2}}=\\frac{y}{-\\left(a^{2}+b^{2}\\right)}=\\frac{1}{a^{2}+b^{2}}$ $\\begin{array}{l|l}\\frac{x}{a^2+b^2}=\\frac{1}{a^2+b^2}&\\frac{y}{-(a^2+b^2)}=\\frac{1}{a^2+b^2}\\\\x=1 & y=-1\\end{array}$& (ii) $\\frac{x}{a}+\\frac{y}{b}=a+b$ $\\frac{x}{a^{2}}+\\frac{y}{b^{2}}=2, a \\neq 0, b \\neq 0$ Sol : $\\frac{x}{a}+\\frac{y}{b}=a+b$ $\\frac{b […]<\/p>\n","protected":false},"author":302,"featured_media":625663,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-625664","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 3.4 - Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 2 \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947: ax + by = a \u2013 bbx \u2013 ay = a + bSol : a1=a, b1=b, c1=a-b a2=b, b2=-a, c2=a+b\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 3.4 - Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923\" \/>\n<meta property=\"og:description\" content=\"Question 2 \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947: ax + by = a \u2013 bbx \u2013 ay = a + bSol : a1=a, b1=b, c1=a-b a2=b, b2=-a, c2=a+b\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-07T12:03:28+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-07T12:03:33+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-4-1-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 3.4 – Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923\",\"datePublished\":\"2023-09-07T12:03:28+00:00\",\"dateModified\":\"2023-09-07T12:03:33+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\"},\"wordCount\":2048,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-4-1-scaled.jpg\",\"articleSection\":[\"Class 10\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/\",\"name\":\"KC Sinha: Exercise 3.4 - 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Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 - IndCareer Schools","description":"Question 2 \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947: ax + by = a \u2013 bbx \u2013 ay = a + bSol : a1=a, b1=b, c1=a-b a2=b, b2=-a, c2=a+b","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 3.4 - Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923","og_description":"Question 2 \u092c\u094d\u0930\u091c-\u0917\u0941\u0923\u0928 \u0935\u093f\u0927\u093f \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923 \u092f\u0941\u0917\u094d\u092e\u094b \u0915\u094b \u0939\u0932 \u0915\u0930\u0947: ax + by = a \u2013 bbx \u2013 ay = a + bSol : a1=a, b1=b, c1=a-b a2=b, b2=-a, c2=a+b","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-07T12:03:28+00:00","article_modified_time":"2023-09-07T12:03:33+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-4-1-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 3.4 – Mathematics Solution Class 10 Chapter 3 \u0926\u094b \u091a\u0930 \u0935\u093e\u0932\u0947 \u0930\u0948\u0916\u093f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923","datePublished":"2023-09-07T12:03:28+00:00","dateModified":"2023-09-07T12:03:33+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/"},"wordCount":2048,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-4-1-scaled.jpg","articleSection":["Class 10"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-4-mathematics-solution-class-10-chapter-3-do-char-wale-raikhik-samikaran\/","name":"KC Sinha: Exercise 3.4 - 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