Solve the system using 2 methods.
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The first method is the substitution method.
We'll write one variable depending on the other one.
From the first equation, we'll choose to write x=6-2y
Now we'll substitute x from the second equation, by it's expression and we'll get an expression in y.
We'll subtract 12 both sides:
Now, we'll substitute y in the x expression.
The second method is to reduce one variable from the both equations of the system.
We'll reduce x and for this reason, we'll multiply the first equation by -2 and we'll get.
Now we'll add the new equation to the second one.
To solve x+2y=6...(1) and 2x+6y=8, using two methods:
From (2) we get: 2x = 8-6y Or x = (8-6y)/2 =4-3y. We substitute x = 4-3y in the equation (1):
4-3y+2y = 6. Or 4-y = 6. Or 4-6 = y. Or y = -2.
Substituting y= -2 in (1), we get: x+2(-2) = 6. Or x= 6+4 = 10.
2y appears in eq (1) and 6y in eq (2). So 3*eq (1)- eq(2) eliminates y:
3*Eq(1) - eq(2): 3(x+2y)-(2x+6y) = 3*6-8 =10.Or
3x-2x = 10. Or x = 10.
Using x = 10, in (2), we get: 10*2 - 6y = 8. Or 6y = 8-20. Or 6y = -12. Or y = -12/6 = -2.
First let us multiply (1) with -2 and add to (2)
==> x= 6-2y = 6-2(-2)= 6+4=10
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