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Sagot :
(-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:)
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:)
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
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
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