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Question

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

3x+2y=11

2x3y+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

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

(224y)33y=10(224y9y)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

=(118)3=33=1

x=1


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