How do you solve this system of equations: #7p + 8q = 11 and 3p + 4q = 3#?

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Answer 1 · 2017-11-26T20:12:48

#p=5\quad,\quad q=-3#

We can use elimination to solve this.

Let’s try to eliminate #q#, to solve for #p# first:

#7p+8q=11#

#3p+4q=3#

Multiply both sides of the bottom equation by #-2#:

#-2(3p+4q)=(3)-2#

#\rightarrow -6p-8q=-6#

Now, the #q# terms in both equations are opposites of each other, so they will cancel out and give us the value of #p#.

To do that, we need to add both equations together:

#(-6p-8q=-6)\quad +\quad (7p+8q=1)#

#\rightarrow p=5#

Knowing that, we can plug the value of #p# into one of the equations to find the value of #q#:

#7p+8q=11#

#\rightarrow 7(5)+8q=11#

#\rightarrow 35+8q=11#

#\rightarrow 8q=-24#

#\rightarrow q=-3#

Now that we have the values of both variables, we can plug them into one of the equations to check our work:

#3p+4q=3#

#3(5)+4(-3)=3#

#15-12=3#

#3=3#

So it’s right.