7x\u2014 15y = 2<\/strong> Solve the following pair of linear equations by substitution method:<\/strong><\/p>\n\n\n\n x + y = 14<\/strong> Solve the following pair of linear equations by substitution method:<\/strong><\/p>\n\n\n\n 3x \u2013 y = 3<\/strong> Solve the following pair of linear equations by substitution method:<\/strong><\/p>\n\n\n\n 0.5x + 0.8y = 3.4<\/strong> Solve the following pair of linear equations by substitution method:<\/strong><\/p>\n\n\n\n x + y = a\u2014b<\/strong> Solve the following pair of linear equations by substitution method:<\/strong><\/p>\n\n\n\n x + y = 2m<\/strong> Solve the following system of equations by substitution method:<\/strong><\/p>\n\n\n\n $\\frac{x}{2}+y=0.8$<\/strong> \u21d2 x + 2y = 0.8\u00d72 Solve the following system of equations by substitution method:<\/strong><\/p>\n\n\n\n s \u2014t = 3<\/strong> Solve the following system of equations by substitution method:<\/strong><\/p>\n\n\n\n $\\frac{x}{a}+\\frac{y}{b}=2$, a \u2260 0, b \u2260 0<\/strong> Solve the following system of equations by substitution method:<\/strong><\/p>\n\n\n\n $\\frac{b x}{a}+\\frac{a y}{b}=a^{2}+b^{2}$<\/strong>
x + 2y = 3<\/strong>
Sol :
Given equations are
7x \u2013 15y = 2 \u2026(i)
x + 2y = 3 \u2026(ii)
From eqn<\/sup> (ii), x = 3 \u2013 2y \u2026(iii)
On substituting x = 3 \u2013 2y in eqn<\/sup> (i), we get
\u21d2 7(3 \u2013 2y) \u2013 15y = 2
\u21d2 21 \u2013 14y \u2013 15y = 2
\u21d2 21 \u2013 29y = 2
\u21d2 \u2013 29y = \u2013 19
$\\Rightarrow \\mathrm{y}=\\frac{19}{29}$
Now, on putting $y=\\frac{19}{29}$ in eqn<\/sup> (iii), we get
$\\Rightarrow x=3-2\\left(\\frac{19}{29}\\right)$
$\\Rightarrow x=3-\\frac{38}{29}$
$\\Rightarrow x=\\frac{87-38}{29}$
$\\Rightarrow x=\\frac{49}{29}$
Thus, $x=\\frac{49}{29}$ and$y=\\frac{19}{29}$ is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 B <\/h4>\n\n\n\n
x \u2013 y = 4<\/strong>
Sol :
Given equations are
x + y = 14 \u2026(i)
x \u2013 y = 4 \u2026(ii)
From eqn<\/sup> (ii), x = 4 + y \u2026(iii)
On substituting x = 4 + y in eqn<\/sup> (i), we get
\u21d2 4 + y + y = 14
\u21d2 4 + 2y = 14
\u21d2 2y = 14 \u2013 4
\u21d2 2y = 10
$\\Rightarrow \\mathrm{y}=\\frac{10}{2}=5$
Now, on putting y = 5 in eqn<\/sup> (iii), we get
\u21d2 x = 4 + 5
\u21d2 x = 9
Thus, x = 9 and![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1543918388446680.png\")
y = 5 is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 <\/h4>\n\n\n\n
9x \u2014 3y = 9<\/strong>
Sol :
Given equations are
3x \u2013 y = 3 \u2026(i)
9x \u2013 3y = 9 \u2026(ii)
From eqn<\/sup> (i), y = 3x \u2013 3 \u2026(iii)
On substituting y = 3x \u2013 3 in eqn<\/sup> (ii), we get
\u21d2 9x \u2013 3(3x \u2013 3) = 9
\u21d2 9x \u2013 9x + 9 = 9
\u21d2 9 = 9
This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 C <\/h4>\n\n\n\n
0.6x \u2014 0.3y = 0.3<\/strong>
Sol :
Given equations are
0.5x + 0.8y = 3.4 \u2026(i)
0.6x \u2013 0.3y = 0.3 \u2026(ii)
From eqn<\/sup> (ii), 2x \u2013 y = 1
y = 2x \u2013 1 \u2026(iii)
On substituting y = 0.2x \u2013 1 in eqn<\/sup> (i), we get
\u21d2 0.5x + 0.8(2x \u2013 1) = 3.4
\u21d2 0.5x + 1.6x \u2013 0.8 = 3.4
\u21d2 2.1x = 3.4 + 0.8
\u21d2 2.1x = 4.2
$\\Rightarrow x=\\frac{4.2}{2.1}=2$
Now, on putting x = 2 in eqn<\/sup> (iii), we get
\u21d2 y = 2(2) \u2013 1
\u21d2 y = 4 \u2013 1
\u21d2 y = 3
Thus, x = 2 and y = 3 is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 A <\/h4>\n\n\n\n
ax \u2014by = a2<\/sup> + b2<\/sup><\/strong>
Sol :
Given equations are
x + y = a \u2013 b \u2026(i)
ax \u2013 by = a2<\/sup> + b2<\/sup> \u2026(ii)
From eqn<\/sup> (i), x = a \u2013 b \u2013 y \u2026(iii)
On substituting x = a \u2013 b \u2013 y in eqn<\/sup> (ii), we get
\u21d2 a(a \u2013 b \u2013 y) \u2013 by = a2<\/sup> + b2<\/sup>
\u21d2 a2<\/sup> \u2013 ab \u2013 ay \u2013 by = a2<\/sup> + b2<\/sup>
\u21d2 \u2013 ab \u2013 y(a + b) = b2<\/sup>
\u21d2 \u2013 y(a + b) = b2<\/sup> + ab
$\\Rightarrow \\mathrm{y}=\\frac{\\mathrm{b}^{2}+\\mathrm{ab}}{-(\\mathrm{a}+\\mathrm{b})}$
$\\Rightarrow \\mathrm{y}=\\frac{\\mathbf{b}(\\mathbf{b}+\\mathbf{a})}{-(\\mathbf{a}+\\mathbf{b})}=-\\mathbf{b}$
Now, on putting y = \u2013 b<\/strong> in eqn<\/sup> (iii), we get
\u21d2 x = a \u2013 b \u2013 ( \u2013 b)
\u21d2 x = a<\/strong>
Thus, x = a<\/strong> and y = \u2013 b<\/strong> is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 B <\/h4>\n\n\n\n
mx \u2014 ny = m2<\/sup> + n2<\/sup><\/strong>
Sol :
Given equations are
x + y = 2m \u2026(i)
mx \u2013 ny = m2<\/sup> + n2<\/sup> \u2026(ii)
From eqn<\/sup> (i), x = 2m \u2013 y \u2026(iii)
On substituting x = 2m \u2013 y in eqn<\/sup> (ii), we get
\u21d2 m(2m \u2013 y) \u2013 ny = m2<\/sup> + n2<\/sup>
\u21d2 2m2<\/sup> \u2013 my \u2013 ny = m2<\/sup> + n2<\/sup>
\u21d2 \u2013 y(m + n) = m2<\/sup> \u2013 2m2<\/sup> + n2<\/sup>
\u21d2 \u2013 y(m + n) = \u2013 m2<\/sup> + n2<\/sup>
$\\Rightarrow y =\\frac{\\mathrm{n}^{2}-\\mathrm{m}^{2}}{-(\\mathrm{a}+\\mathrm{b})}\\left(\\because\\left(\\mathrm{a}^{2}-\\mathrm{b}^{2}\\right)=(\\mathrm{a}-\\mathrm{b})(\\mathrm{a}+\\mathrm{b})\\right)$
$\\Rightarrow \\mathrm{y}=\\frac{(\\mathrm{n}-\\mathrm{m})(\\mathrm{n}+\\mathrm{m})}{-(\\mathrm{n}+\\mathrm{m})}$
\u21d2 y = \u2013 (n \u2013 m) = m \u2013 n
Now, on putting y = m \u2013 n<\/strong> in eqn<\/sup> (iii), we get
\u21d2 x = 2m \u2013 (m \u2013 n)
\u21d2 x = 2m \u2013 m + n
\u21d2 x = m + n<\/strong>
Thus, x = m + n<\/strong> and y = m \u2013 n<\/strong> is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 A <\/h4>\n\n\n\n
$x+\\frac{y}{2}=\\frac{7}{10}$<\/strong>
Sol :
Given equations are
$\\frac{x}{2}+y=0.8$ \u2026(i)
$x+\\frac{y}{2}=\\frac{7}{10}$ \u2026(ii)
eqn<\/sup> (i) can be re \u2013 written as,
$\\Rightarrow \\frac{x}{2}+y=0.8$
$\\Rightarrow\\left(\\frac{x+2 y}{2}\\right)=0.8$<\/p>\n\n\n\n
\u21d2 x + 2y = 1.6 \u2026(iii)
On substituting x = 1.6 \u2013 2y in eqn<\/sup> (ii), we get
$\\Rightarrow 1.6-2 y+\\frac{y}{2}=\\frac{7}{10}$
$\\Rightarrow \\frac{3.2-4 y+y}{2}=\\frac{7}{10}$
$\\Rightarrow 3.2-3 \\mathrm{y}=\\frac{7}{10} \\times 2$
$\\Rightarrow-3 y=\\frac{7}{10} \\times 2-3.2$
$\\Rightarrow-3 y=\\frac{14}{10}-\\frac{32}{10}$
$\\Rightarrow-3 y=-\\frac{18}{10}$
$\\Rightarrow \\mathrm{y}=\\frac{18}{10} \\times \\frac{1}{3}$
$\\Rightarrow \\mathrm{y}=\\frac{6}{10}=0.6$
Now, putting the y = 0.6 in eqn<\/sup> (iii), we get
\u21d2 x + 2y = 1.6
\u21d2 x + 2(0.6) = 1.6
\u21d2 x + 1.2 = 1.6
\u21d2<\/strong> x = 0.4<\/strong>
Thus, x = 0.4<\/strong> and y = 0.6<\/strong> is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 B <\/h4>\n\n\n\n
$\\frac{\\mathrm{s}}{3}+\\frac{\\mathrm{t}}{2}=6$<\/strong>
Sol :
Given equations are
s \u2013 t = 3 \u2026(i)
$\\frac{s}{3}+\\frac{t}{2}=6$ \u2026(ii)
From eqn<\/sup> (i), we get
\u21d2 s = 3 + t \u2026(iii)
On substituting s = 3 + t in eqn<\/sup> (ii), we get
$\\Rightarrow \\frac{3+t}{3}+\\frac{t}{2}=6$
$\\Rightarrow \\frac{2(3+t)+3 t}{6}=6$
\u21d2 6 + 2t + 3t = 6\u00d76
\u21d2 5t = 36 \u2013 6
$\\Rightarrow \\mathrm{t}=\\frac{30}{5}=6$
Now, putting the t = 6 in eqn<\/sup> (iii), we get
\u21d2 s = 3 + 6
\u21d2<\/strong> s = 9<\/strong>
Thus, s = 9<\/strong> and![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/154391840573997.png\")
t = 6<\/strong> is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 C <\/h4>\n\n\n\n
ax \u2013 by = a2<\/sup> \u2013 b2<\/sup><\/strong>
Sol :
Given equations are
$\\frac{x}{a}+\\frac{y}{b}=2$ \u2026(i)
ax \u2013 by = a2<\/sup> \u2013 b2<\/sup> \u2026(ii)
eqn<\/sup> (i) can be re \u2013 written as,
$\\Rightarrow \\frac{x}{a}+\\frac{y}{b}=2$
$\\Rightarrow\\left(\\frac{b x+a y}{a b}\\right)=2$
\u21d2 bx + ay = ab\u00d72
\u21d2 bx + ay = 2ab
$\\Rightarrow x=\\frac{2 a b-a y}{b}$ \u2026(iii)
On substituting $\\mathrm{X}=\\frac{2 \\mathrm{ab}-\\mathrm{ay}}{\\mathrm{b}}$ in eqn<\/sup> (ii), we get
$\\Rightarrow \\mathrm{a}\\left(\\frac{2 \\mathrm{ab}-\\mathrm{ay}}{\\mathrm{b}}\\right)-\\mathrm{by}=\\mathrm{a}^{2}-\\mathrm{b}^{2}$
$\\Rightarrow \\frac{2 a^{2} b-a^{2} y}{b}-b y=a^{2}-b^{2}$
$\\Rightarrow \\frac{2 a^{2} b-a^{2} y-b^{2} y}{b}=a^{2}-b^{2}$
\u21d2 2a2<\/sup> b \u2013 a2<\/sup> y \u2013 b2<\/sup> y = b(a2<\/sup> \u2013 b2<\/sup>)
\u21d2 2a2<\/sup> b \u2013 y(a2<\/sup> + b2<\/sup> ) = a2<\/sup>b \u2013 b3<\/sup>
\u21d2<\/strong> \u2013 y(a2<\/sup> + b2<\/sup> ) = a2<\/sup> b \u2013 b3<\/sup> \u2013 2a2<\/sup>b
\u21d2<\/strong> \u2013 y(a2<\/sup> + b2<\/sup> ) = \u2013 a2<\/sup>b \u2013 b3<\/sup>
\u21d2<\/strong> \u2013 y(a2<\/sup> + b2<\/sup> ) = \u2013 b(a2<\/sup> + b2<\/sup>)
$\\Rightarrow y=\\frac{-b\\left(a^{2}+b^{2}\\right)}{-\\left(a^{2}+b^{2}\\right)}=b$
Now, on putting y = b<\/strong> in eqn<\/sup> (iii), we get
$\\Rightarrow x=\\frac{2 a b-a b}{b}$
\u21d2 x = a<\/strong>
Thus, x = a<\/strong> and y = b<\/strong> is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 D <\/h4>\n\n\n\n
x + y = 2ab<\/strong>
Sol :
Given equations are
$\\frac{b x}{a}+\\frac{a y}{b}=a^{2}+b^{2}$ \u2026(i)
x + y = 2ab \u2026(ii)
eqn<\/sup> (i) can be re – written as,
$\\Rightarrow \\frac{b x}{a}+\\frac{a y}{b}=a^{2}+b^{2}$
$\\Rightarrow\\left(\\frac{\\mathrm{b}^{2} \\mathrm{x}+\\mathrm{a}^{2} \\mathrm{y}}{\\mathrm{ab}}\\right)=\\mathrm{a}^{2}+\\mathrm{b}^{2}$
\u21d2 b2<\/sup> x + a2<\/sup> y = ab \u00d7 (a2<\/sup> + b2<\/sup>)
$\\Rightarrow x=\\frac{a b \\times\\left(a^{2}+b^{2}\\right)-a^{2} y}{b^{2}}$ \u2026(iii)
Now, from eqn<\/sup> (ii), y = 2ab \u2013 x \u2026(iv)
On substituting y = 2ab \u2013 x in eqn<\/sup> (iii), we get
$\\Rightarrow x=\\frac{a b \\times\\left(a^{2}+b^{2}\\right)-a^{2}(2 a b-x)}{b^{2}}$
$\\Rightarrow x=\\frac{a^{3} b+b^{3} a-2 a^{3} b+a^{2} x}{b^{2}}$
\u21d2 b2<\/sup> x = b3<\/sup> a \u2013 a3<\/sup>b + a2<\/sup>x
\u21d2 b2<\/sup>x \u2013 a2<\/sup>x = b3<\/sup>a \u2013 a3<\/sup>b
\u21d2 (b2 \u2013<\/sup> a2<\/sup>) x = ab(b2<\/sup> \u2013 a2<\/sup>)
\u21d2<\/strong> x = ab<\/strong>
Now, on putting x = ab<\/strong> in eqn<\/sup> (iv), we get
\u21d2 y = 2ab \u2013 ab
\u21d2 y = ab<\/strong>
Thus, x = ab<\/strong> and y = ab<\/strong> is the required solution.<\/p>\n\n\n\n