[5 marks sum]

Question 515 Marks

Solve the following pair of linear $($Simultaneous$)$ equation using method of elimination by substitution :$2( x - 3 ) + 3( y - 5 ) = 0;5( x - 1 ) + 4( y - 4 ) = 0$

Answer

Given equations are
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3)$, we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.

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Question 525 Marks

Solve the following pair of linear $($simultaneous$)$ equation by the method of elimination by substitution:$1.5x + 0.1y = 6.2;3x - 0.4y = 11.2$

Answer

The given pair of linear equations are
$1.5 x+0.1 y=6.2\ldots(1)$
$3 x-0.4 y=11.2\ldots(2)$
equation $(1)$ dividing from $10$
$\frac{1.5 x}{10}+\frac{0.1 y}{10}=\frac{6.2}{10}$
$ \Rightarrow 15 x+y=62$
$y=62-15 x\ldots(3)$
Putting the value of $y$ from in equation $(2)$
$3 x-0.4(62-15 x)=11.2$
$ \Rightarrow \frac{30 x-4(62-15 x)}{10}=\frac{112}{10}$
$ \Rightarrow 30 x-248+60 x=112$
$ \Rightarrow 90 x-248=112$
$ \Rightarrow 90 x=360$
$\Rightarrow x=4\dots ....(4)$
Substitute the value of $x$ from equation $(4)$ in equation $(3)$
$y=62-15 x$
$ =62-15(4)$
$ =62-60$
$ \Rightarrow y=2$
$\therefore$ Solution is $\mathrm{x}=4$ and $\mathrm{y}=2$.

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Question 535 Marks

Solve the following pair of linear $($simultaneous$)$ equation by the method of elimination by substitution:$0.2x + 0.1y = 25;2(x - 2) - 1.6y = 116$

Answer

The given pair of linear equations are
$0.2 x+0.1 y=25\ldots .(1)$
$2(x-2)-1.6 y=116\dots ..(2)$
Consider equation $(1)$
$0.2 x+0.1 y=25$
$ \Rightarrow 0.2 x=25-0.1 y$
$\Rightarrow x=\frac{25-0.1 y}{0.2}\ldots .(3)$
Substitute the value of $x$ from equation $(3)$ in equation $(2).$
$2(x-2)-1.6 y=116$
$ \Rightarrow 2\left[\frac{25-0.1 y}{0.2}-2\right]-1.6 y=116$
$ \Rightarrow 2 \times \frac{25-0.1}{0.2}-2 \times 2-1.6 y=116$
$ \Rightarrow 0.2 \times \frac{50-0.2}{0.2}-4-1.6 y=116$
$ \Rightarrow 50-0.2 y-0.8-0.32 y=23.2$
$ \Rightarrow 26=0.52 y$
$\Rightarrow y=50 \dots....(4)$
Substitute the value of $y$ from equation $(4)$ in equation $(3).$
$x=\frac{25-0.1 y}{0.2}$
$ \Rightarrow x=\frac{25-0.1(50)}{0.2}$
$ \Rightarrow x=\frac{25-5}{0.2}$
$ \Rightarrow x=100$
$\therefore$ Solution is $x=100$ and $y=50$.

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Question 545 Marks

Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution :$8x + 5y = 9,3x + 2y = 4$

Answer

$8 x+5 y=9 \dots (1)$
$3 x+2 y=4 \ldots $
$ 8 x+5 y=9$
$ \therefore 5 y=9-8 x $
$ \therefore y=\frac{9-8 x}{5} \ldots .$
Putting this value of $y$ in $(2)$
$ 3 x+2\left[\frac{9-8 x}{5}\right]=4 $
Multiplying by $5,$
$15 x+18-16 x=20 $
$ 15 x-16 x=20-18$
$-x=2$
$ x=-2$
From $(3)\ y=\frac{9-8 x}{5}=\frac{9-8(-2)}
{5}=\frac{25}{5}=5$
$ y=5 $

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Question 555 Marks

Solve the following pairs of linear $($simultaneous$)$ equation using method of elimination by substitution:$\frac{x}{6}+\frac{y}{15}=4;\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$

Answer

$ \frac{x}{6}+\frac{y}{15}=4$
$ \Rightarrow \frac{5 x+2 y}{30}=4$
$ \Rightarrow 5 x+2 y=120$
$ \Rightarrow 5 x=120-2 y$
$\Rightarrow x=\frac{120-2 y}{5}$$\ldots .(1)$
And,
$\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
$\Rightarrow \frac{1}{3}\left(x-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{1}{3}\left(\frac{120-2 y}{5}-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{480-8 y-5 y}{20}=\frac{57}{4}$
$\Rightarrow \frac{480-13 y}{20}=\frac{57}{4}$
$\Rightarrow 480-13 y=285$
$\Rightarrow 13 y=195$
$ \Rightarrow y=15$
Substituting the value of $y$ in $(1)$, we have
$x=\frac{120-2 \times 15}{5}=\frac{120-30}{5}=\frac{90}{5}=18$
$\therefore$ Solution is $x=18$ and $y=15$

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Question 565 Marks

Solve the following pair of linear $($simultaneous$)$ equation using method of elimination by substitution:$\frac{3 x}{2}-\frac{5 y}{3}+2=0,\frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$

Answer

$ \frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$
$ \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$
$ \Rightarrow \frac{2 x+3 y}{6}=\frac{13}{6}$
$ \Rightarrow 2 x+3 y=13$
$ \Rightarrow 2 x=13-3 y$
$\Rightarrow x=\frac{13-3 y}{2}\ldots .(1)$
And,
$\frac{3 x}{2}-\frac{5 y}{3}+2=0$
$ \Rightarrow \frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{39-9 y}{4}-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{117-27 y-20 y}{12}=-2$
$ \Rightarrow \frac{117-47 y}{12}=-2$
$ \Rightarrow 117-47 y=-24$
$ \Rightarrow 47 y=141$
$ \Rightarrow y=3$
Substituting the value of $y$ in $(1)$, we have
$x=\frac{13-3 \times 3}{2}=\frac{13-9}{2}=\frac{4}{2}=2$
$\therefore$ Solution is $x=2$ and $y=3$.

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Question 575 Marks

Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12;\frac{3 x+2}{2}-\frac{4 y+3}{9}=13$

Answer

$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12 \quad ($given$)$
$ \Rightarrow \frac{3(2 x+1)+7(5 y-3)}{21}=12$
$ \Rightarrow 6 x+3+35 y-21=252$
$ \Rightarrow 6 x+35 y-18=252$
$ \Rightarrow 6 x+35 y=270$
$ \Rightarrow 6 x=270-35 y$
$ \Rightarrow x=\frac{270-35 y}{6}$
$\frac{3 x+2}{2}-\frac{4 y+3}{9}=13 \quad ($given$)$
$ \Rightarrow \frac{9(3 x+2)-2(4 y+3)}{18}=13$
$ \Rightarrow 27 x+18-8 y-6=234$
$ \Rightarrow 27 x-8 y+12=234$
$\Rightarrow 27 x-8 y=222\ldots .(1)$
Substituting $x=\frac{270-35 y}{6}$ in $(1),$ we get
$27\left(\frac{270-35 y}{6}\right)-8 y=222$
$ \Rightarrow 7290-945 y-48 y=1332$
$ \Rightarrow-993 y=-5958$
$ \Rightarrow y=6$
Substituting $y=6$ in $x=\frac{270-35 y}{6}$, we get
$x=\frac{270-35 \times 6}{6}=\frac{270-210}{6}=\frac{60}{6}=10$
$\therefore$ Solution is $x=10$ and $y=6$.

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MATHEMATICS [5 marks sum]