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

3 Answers
Aug 13, 2017

#(x,y)=(8,-6)#

Explanation:

Given
[1]#color(white)("XXX")y=(-3)/4x#
[2]#color(white)("XXX")x-4y=32#

Using [1] we can substitute #(-3)/4x# for #y# in [2] to get
[3]#color(white)("XXX")x-4xx(-3)/4x=32#

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

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

Dividing both sides of [6] by #4#
[7]#color(white)("XXX")x=8#

Using [7] we can substitute #8# for #x# in [1] to get
[8]#color(white)("XXX")y=(-3)/4xx 8#

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

Aug 13, 2017

#(x,y)to(8,-6)#

Explanation:

#color(red)(y)=-3/4xto(1)#

#x-4color(red)(y)=32to(2)#

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

#rArrx-(4xx-3/4)=32#

#rArrx+3x=32#

#rArr4x=32#

#"divide both sides by 4"#

#rArrx=8#

#"substitute this value in "(1)#

#y=-3/4xx8=-6#

#color(blue)"As a check"#

#"substitute these values in "(2)#

#8+24=32larr" True"#

#rArr"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]}

Aug 13, 2017

Substitution.

#x=8#
#y=-6#

Explanation:

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

This is Eq 2:

#x-4color(green)y=32#

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

#x-4(-3/4x)=32#
#x+3x=32#
#4x=32#
#x=8#

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

#y=-3/4x#
#y=-3/4(8)=-6#

So, the solution to the system of equations is:

#x=8#
#y=-6#

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

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