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

$2 x + 7 y = 39$

$3 x + 5 y = 31$

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Solution

2x + 7y = 39...(1)

3x + 5y = 31...(2)
From (1), x= $open square brackets fraction numerator 39 minus 7 y over denominator 2 end fraction close square brackets$

Putting this value of x in (2)

3 $open square brackets fraction numerator 39 minus 7 y over denominator 2 end fraction close square brackets$ +5y=31

117 - 21y + 10y = 62

-11y = -55

y = 5

From (1) x= $open square brackets fraction numerator 39 minus 7 left parenthesis 5 right parenthesis over denominator 2 end fraction close square brackets$

x=4/2
x=2

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Similar questions

2 x + 7 y = 39

3 x + 5 y = 31

2 x + 7 y = 39

3 x + 5 y = 31

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

$2 x - 3 y = 7$

$5 x + y = 9$

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