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

$1.5 x + 0.1 y = 6.2$

$3 x - 0.4 y = 11.2$

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Solution

The given pair of linear equations are

1.5x0.1y=6.2

3x-0.4y=11.2

1.5x+0.1y=6.2

1.5x=6.2-0.1y

x= $open square brackets fraction numerator 6.2 minus 0.1 y over denominator 1.5 end fraction close square brackets$

3x-0.4y=11.2

3 $open square brackets fraction numerator 6.2 minus 0.1 y over denominator 1.5 end fraction close square brackets$ -0.4y=11.2

2(6.2-0.1y)-0.4y=11.2

-0.6y=-1.2

y=2

x= $open square brackets fraction numerator 6.2 minus 0.1 y over denominator 1.5 end fraction close square brackets$

x= $fraction numerator 6.2 minus 0.1 left parenthesis 2 right parenthesis over denominator 1.5 end fraction$

x= $open square brackets fraction numerator 6.2 minus 0.2 over denominator 1.5 end fraction close square brackets$

x=4 & y=2

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Similar questions

1.5 x + 0.1 y = 6.2

3 x - 0.4 y = 11.2

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$6 x = 7 y + 7$

$7 y - x = 8$

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2 (x - 2) - 1.6 y = 116

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