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There's bound to be only one pair of values that satisfies <i>both</i> equations simultaneously, which is what the &quot;substitution method&quot; is meant to find.
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There's bound to be only one pair of values that satisfies both equations simultaneously, which is what the "substitution method" is meant to find.

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CPColin, you are correct. I thought we were looking at two separate equations. There will probably be one or two values which will satisfy both equations. I will work out those values and post the solution shortly.
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CPColin, you are correct. I thought we were looking at two separate equations. There will probably be one or two values which will satisfy both equations. I will work out those values and post the solution shortly.

0.05x-0.01y=1.1
0.27x+0.14y=13.7
5x-1y=110
5x-110=y
0.27x+0.14(5x-110)=13.7
27x+14(5x-110)=1370
27x+70x-1540=1370
97x=2910
x=30
y-5(30)-110
y=40

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ozo beat me to it. His answer is correct.
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ozo beat me to it. His answer is correct.

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0.05x-0.01y=1.1<br />
0.27x+0.14y=13.7<br />
Need a solution set,empty set, or infinite.
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0.05x-0.01y=1.1
0.27x+0.14y=13.7
Need a solution set,empty set, or infinite.

solve the system of equations by the substitution method

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