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Sagot :
15x + 15y = 60
5x - 5y = 40
To use elimination by addition, you need to get variables in one equation the same but negative in the other equation. For example, to use addition elimination to match the first equation, the second equation has to be - 15x or -15y.
Luckily, the second equation is easily manipulated to get to -15y.
To go from -5y to -15y you just need to multiply by 3, so we multiply the whole second equation by 3:
3(5x - 5y = 40)
15x - 15y = 120
Now we can add the two equations to "eliminate" the y's
15x + 15 y = 60
+ 15x - 15y = 120
30x + 0y = 180
now solve for x:
30x = 180
divide both sides by 30
x = 6
Now that we solved for x, we can plug it back into the original equations to get y:
15(6) + 15y = 60
90 + 15y = 60
15y = -30
y = -2
5(6) - 5y = 40
30 - 5y = 40
-5y = 10
y = -2
x = 6 and y = -2
5x - 5y = 40
To use elimination by addition, you need to get variables in one equation the same but negative in the other equation. For example, to use addition elimination to match the first equation, the second equation has to be - 15x or -15y.
Luckily, the second equation is easily manipulated to get to -15y.
To go from -5y to -15y you just need to multiply by 3, so we multiply the whole second equation by 3:
3(5x - 5y = 40)
15x - 15y = 120
Now we can add the two equations to "eliminate" the y's
15x + 15 y = 60
+ 15x - 15y = 120
30x + 0y = 180
now solve for x:
30x = 180
divide both sides by 30
x = 6
Now that we solved for x, we can plug it back into the original equations to get y:
15(6) + 15y = 60
90 + 15y = 60
15y = -30
y = -2
5(6) - 5y = 40
30 - 5y = 40
-5y = 10
y = -2
x = 6 and y = -2
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