Solve the simultaneous equations:

\[ 4x + 7y = 1 \]

\[ 3x + 10y = 15 \]

(-3)× (4x+7y=1) -12x-21y=-3 4× (3x+10y=15) 12x+40y=60 ⇒ (x s cancel out)
-21y=-3 40y=60 ⇒19y-57 y=3 for find the x plug in the 3 to the y one of these two equations and solve for the x
12x+40(3)=60 12x=-60 **x=5 ** HOPE IT HELPS:)

Answered by PrudentAysegul | 2024-06-10

We are going to be adding the two equations and want to cancel something out to isolate the other. Let's cancel out the x term first Multiply the first equation by -3 and the second equation by 4 -3(4x+7y)=-3(1) => -12x-21y=-3 4(3x+10y)=4(15) => 12x+40y=60 Add the two equations and the x terms cancel out, leaving 19y=57 y=57/9=3 Substitute that into the second equation and you get 3x+10(3)=15 3x+30=15 3x=-15 X=-15/3=-5 Final answer: x=-5 y=3

Answered by FirstSineOfMadness | 2024-06-10

The solution to the simultaneous equations 4 x + 7 y = 1 and 3 x + 10 y = 15 is x = − 5 and y = 3 . This is found using the method of elimination. The derived values satisfy both original equations.
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Answered by PrudentAysegul | 2024-10-02