How do you solve this system of equations y= \frac { - 3} { 4} x , x - 4y = 32y=−34x,x−4y=32?

3 Answers
Alan P.
Aug 13, 2017

(x,y)=(8,-6)(x,y)=(8,−6)

Explanation:

Given
[1]color(white)("XXX")y=(-3)/4xXXXy=−34x
[2]color(white)("XXX")x-4y=32XXXx−4y=32

Using [1] we can substitute (-3)/4x−34x for yy in [2] to get
[3]color(white)("XXX")x-4xx(-3)/4x=32XXXx−4×−34x=32

Simplifying [4]
[5]color(white)("XXX")x-(-3)x=32XXXx−(−3)x=32

Continuing the simplification:
[6]color(white)("XXX")4x=32XXX4x=32

Dividing both sides of [6] by 44
[7]color(white)("XXX")x=8XXXx=8

Using [7] we can substitute 88 for xx in [1] to get
[8]color(white)("XXX")y=(-3)/4xx 8XXXy=−34×8

Simplifying [8]
[9]color(white)("XXX")y=-6XXXy=−6

Jim G.
Aug 13, 2017

(x,y)to(8,-6)(x,y)→(8,−6)

Explanation:

color(red)(y)=-3/4xto(1)y=−34x→(1)

x-4color(red)(y)=32to(2)x−4y=32→(2)

"substitute "y=-3/4x" in "(2)substitute y=−34x in (2)

rArrx-(4xx-3/4)=32⇒x−(4×−34)=32

rArrx+3x=32⇒x+3x=32

rArr4x=32⇒4x=32

"divide both sides by 4"divide both sides by 4

rArrx=8⇒x=8

"substitute this value in "(1)substitute this value in (1)

y=-3/4xx8=-6y=−34×8=−6

color(blue)"As a check"As a check

"substitute these values in "(2)substitute these values in (2)

8+24=32larr" True"8+24=32← True

rArr"point of intersection "=(8,-6)⇒point of intersection =(8,−6) graph{(y+3/4x)(y-1/4x+8)((x-8)^2+(y+6)^2-0.06)=0 [-12.49, 12.48, -6.25, 6.24]}

Sally
Aug 13, 2017

Substitution.

x=8x=8
y=-6y=−6

Explanation:

There are many ways to solve systems of equations. For this system:
color(green)y=-3/4xy=−34x (Eq. 1)
x-4y=32x−4y=32 (Eq. 2),
it would be easiest to solve it with substitution since Equation (Eq.) 1 is already solved for yy. This means we can simply plug in the yy value in the second equation.

This is Eq 2:

x-4color(green)y=32x−4y=32

If we plug in Eq. 1 into Eq. 2, we get:

x-4(-3/4x)=32x−4(−34x)=32
x+3x=32x+3x=32
4x=324x=32
x=8x=8

Now we solved for the first variable. To solve for yy, all we do is plug in our value of xx back into Eq. 1:

y=-3/4xy=−34x
y=-3/4(8)=-6y=−34(8)=−6

So, the solution to the system of equations is:

x=8x=8
y=-6y=−6

To check this answer, you can plug in the xx and yy values into Eq.1 and Eq. 2 respectively to see if the equation solves correctly:

Eq 1 verification by plugging in the xx and yy values:
y=-3/4xy=−34x
-6=-3/4(8)=-6−6=−34(8)=−6 (so it works)
Eq 2 verification by plugging in the xx and yy values:
x-4y=32x−4y=32
8-4(-6)=8+24=328−4(−6)=8+24=32 (so it works)

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