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
{ 3 x + 10 y = 15 ∣ ∗ 4 4 x + 7 y = 1 ∣ ∗ ( − 3 ) { 12 x + 40 y = 60 − 12 x − 21 y = − 3 + − − − − − A dd i t i o n m e t h o d − 21 y + 40 y = − 3 + 60 19 y = 57 ∣ : 19 y = 3 4 x = 1 − 7 y x = 4 1 − 7 y x = 4 1 − 7 ∗ 3 = 4 − 20 = − 5 S o l u t i o n : x = − 5 y = 3
To solve the simultaneous equations 4 x + 7 y = 1 and 3 x + 10 y = 15 , we find that y = 3 and x = − 5 . The elimination method was used, ensuring that both equations lined up correctly. Finally, we confirmed the solution by substituting back into the original equations.
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