Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

$3 x + 2 y = 11$

$2 x - 3 y + 10 = 0$

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Solution

3x+2y=11

3x=11-2y

x = $open square brackets fraction numerator 11 minus 2 y over denominator 3 end fraction close square brackets$

2x-3y+10=0

2 $open square brackets fraction numerator 11 minus 2 y over denominator 3 end fraction close square brackets$ -3y+10=0

$\frac{(22 - 4 y)}{3} - 3 y = - 10 \frac{(22 - 4 y - 9 y)}{3} = - 10$

22-13y=-30

13y=52

y=4

x= $open square brackets fraction numerator 11 minus 2 left parenthesis 4 right parenthesis over denominator 3 end fraction close square brackets$

$= \frac{(11 - 8)}{3} = \frac{3}{3} = 1$

x=1

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