Question

Solve the following simultaneous equations.
(1) 3a + 5b = 26; a + 5b = 22
(2) x + 7y = 10; 3x – 2y = 7
(3) 2x – 3y = 9; 2x + y = 13
(4) 5m – 3n = 19; m – 6n = –7
(5) 5x + 2y = –3; x + 5y = 4
(6) $\frac{1}{3}x+y=\frac{10}{3};2x+\frac{1}{4}y=\frac{11}{4}$
(7) 99x + 101y = 499; 101x + 99y = 501
(8) 49x – 57y = 172; 57x – 49y = 252

Answer 1

(1) 3a + 5b = 26; .....(I)
a + 5b = 22 .....(II)
Subtracting (II) from (I)
2a = 4
$\Rightarrow$ a = 2
Putting the value of a = 2 in (II)
5b = 22 $-$ 2 = 20
$\Rightarrow$b = $\frac{20}{5}=4$
Thus, a = 2 and b = 4.

(2) x + 7y = 10; .....(I)
3x – 2y = 7 .....(II)
Multiplying (I) with 3
3x + 21y = 30; .....(III)
3x – 2y = 7 .....(IV)
Subtracting (IV) from (III) we get
23y = 23
$\Rightarrow$y = 1
Putting the value of y in (IV) we get
3x – 2 = 7
$\Rightarrow$3x = 7 + 2 = 9
$\Rightarrow$3x = 9
$\Rightarrow$x = 3
Thus, (x, y) = (3, 1)

(3) 2x – 3y = 9 .....(I)
2x + y = 13 .....(II)
Subtracting (II) from (I) we get
– 3y − y = 9 − 13
$$\begin{gathered} \Rightarrow -4y=-4 \\ \Rightarrow y=1 \end{gathered}$$
Putting this value in (I) we get
$$\begin{gathered} 2x-3\left(1\right)=9 \\ \Rightarrow 2x=9+3=12 \\ \Rightarrow x=\frac{12}{2}=6 \end{gathered}$$
Thus, (x, y) = (6, 1)

(4) 5m – 3n = 19 .....(I)
m – 6n = –7 .....(II)
Multiplying (I) with 2 we get
10m – 6n = 38 .....(III)
m – 6n = –7 .....(IV)
Subtracting (IV) from (III) we get
$$\begin{gathered} 10m-m-6n-\left(-6n\right)=38-\left(-7\right) \\ \Rightarrow 9m=45 \\ \Rightarrow m=\frac{45}{9}=5 \end{gathered}$$
Putting the value of m = 5 in (II) we get
$$\begin{gathered} 5-6n=-7 \\ \Rightarrow -6n=-7-5 \\ \Rightarrow -6n=-12 \\ \Rightarrow n=\frac{-12}{-6}=2 \end{gathered}$$
Thus, (m, n) = (5, 2).

(5) 5x + 2y = –3 .....(I)
x + 5y = 4 .....(II)
Multiply (II) with 5 we get
5x + 25y = 20 .....(III)
Subtracting (III) from (I) we get
$$\begin{gathered} 5x-5x+2y-25y=-3-20 \\ \Rightarrow -23y=-23 \\ \Rightarrow y=\frac{-23}{-23}=1 \end{gathered}$$
Putting the value of y = 1 in (II) we get
$$\begin{gathered} x+5\left(1\right)=4 \\ \Rightarrow x+5=4 \\ \Rightarrow x=4-5=-1 \end{gathered}$$
Thus, (x, y) = (−1, 1)

(6)
$$\begin{gathered} \frac{1}{3}x+y=\frac{10}{3}.....\left(\mathrm{I}\right) \\ 2x+\frac{1}{4}y=\frac{11}{4}.....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 3 and (II) with 4
$$\begin{gathered} x+3y=10.....\left(\mathrm{III}\right) \\ 8x+y=11.....\left(\mathrm{IV}\right) \end{gathered}$$
Multiply (IV) with 3
24x + 3y = 33 .....(V)
Subtracting (V) from (III)
$$\begin{gathered} x-24x+3y-3y=10-33 \\ \Rightarrow -23x=-23 \\ \Rightarrow x=1 \end{gathered}$$
Putting the value of x = 1 in (III)
$$\begin{gathered} 1+3y=10 \\ \Rightarrow 3y=10-1=9 \\ \Rightarrow y=\frac{9}{3}=3 \end{gathered}$$
Thus, (x, y) = (1, 3)

(7) 99x + 101y = 499 .....(I)
101x + 99y = 501 .....(II)
Adding (I) and (II)
$$\begin{gathered} 200x+200y=1000 \\ \Rightarrow x+y=5.....\left(\mathrm{III}\right) \end{gathered}$$
Subtracting (II) from (I)
$$\begin{gathered} 99x-101x+101y-99y=499-501 \\ \Rightarrow -2x+2y=-2 \\ \Rightarrow -x+y=-1.....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV)
$$\begin{gathered} x+y=5 \\ -x+y=-1 \\ \Rightarrow 2y=4 \\ \Rightarrow y=2 \end{gathered}$$
Putting the value of y = 2 in (III) we get
$$\begin{gathered} x+2=5 \\ \Rightarrow x=5-2=3 \end{gathered}$$
Thus, (x, y) = (3, 2)

(8) 49x – 57y = 172 .....(I)
57x – 49y = 252 .....(II)
Adding (I) and (II)
$$\begin{gathered} 49x+57x-57y-49y=172+252 \\ \Rightarrow 106x-106y=424 \\ \Rightarrow x-y=4.....\left(\mathrm{III}\right) \end{gathered}$$
Subtracting (II) from (I) we have
$$\begin{gathered} 49x-57y-57y-\left(-49y\right)=252-172 \\ \Rightarrow -8x-8y=-80 \\ \Rightarrow -x-y=-10 \\ \Rightarrow x+y=10.....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV)

$$\begin{gathered} x-y=4 \\ x+y=10 \\ \Rightarrow 2x=14 \\ \Rightarrow x=7 \end{gathered}$$
Putting the value of x = 7 in (IV) we get
$$\begin{gathered} 7+y=10 \\ \Rightarrow y=10-7 \\ \Rightarrow y=3 \end{gathered}$$
Thus, (x, y) = (7, 3).