# solve the linear system 4x-y+12=0 3x-y+6=0

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### 3 Answers

We'll re-write the equations of the system, isolating the variable y to the left side:

The equations are:

y = 4x + 12 (1)

y = 3x + 6 (2)

We'll put (1) = (2)

4x + 12 = 3x + 6

We'll isolate x to the left side. For this reason, we'll subtract 3x both sides:

4x - 3x + 12 = 6

x + 12 = 6

We'll subtract 12 both sides:

x = -12 + 6

x = -6

We'll substitute x in (2):

y = 3*(-6) + 6

y = -18 + 6

y = -12

**The solution of the system is: (-6 , -12).**

4x-y+12 = 3x-y+6

4x - 3x - y + y = 6 - 12

simplify

x = -6

now plug in -6 as x

4(-6) - y + 12 = 0

-24 - y + 12 = 0

-24 + 12 = y

-12 = y

(-6,-12)

4x-y+12 = 3x-y+6

get like terms on the same side

4x - 3x - y + y = 6 - 12

simplify

**x = -6**

now plug in -6 as x into any one of the problems to solve for y:

4(-6) - y + 12 = 0

-24 - y + 12 = 0

add y to both sides:

-24 + 12 = y

combine like terms

**-12 = y**

(-6,-12)