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Dafna11 [192]
3 years ago
10

Solve the simultaneous equations: 4x+7y=1 3x+10y=15

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0
Ok we need a common coefficient of either x or y:
12x+21y=3
12x+40y=60
We can subtract one from the other to get:
19y=57
Now divide by 19 to get:
y=3
Now if we sub this back in we see that:
3x+30=15
3x=-15
x=-5
trasher [3.6K]3 years ago
6 0
\left \{ {{4x+7y=1\ \ \ |*(-3)} \atop {3x+10y=15\ \ |*4}} \right. \\\\ \left \{ {{-12x-21y=-3} \atop {12x+40y=60}} \right. \\+-----\\Addition\ method\\\\\ -21y+40y=-3+60\\19y=57\ \ |:19\\\\y=3\\4x=1-7y\\x=\frac{1-7y}{4}\\x=\frac{1-7*3}{4}=\frac{-20}{4}=-5\\\\
Solution:\\
x=-5\\y=3

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