Question #bbb03

2 Answers
Dec 15, 2017

You solve this as a systems of equations.

Explanation:

Start by multiplying the entire equation of #y=4x-7# by #7# and the entire equation of #y=-7x+15# by #4#. Now you have:

#7y=28x-49" "# and #" "4y=-28x+60#

Line them up vertically:

#7y=28x-49#
#4y=-28x+60#

and subtract them, so the #x#'s cancel out. You now should have

#3y=11#

Divide both sides by #3# to get #y# by itself, and you get

#y = 11/3. #

Now, plug in #11/3# for #y# into the original equation and solve for #x#:

#11/3=4x-7" "# (add #7# to both sides to cancel it)

#32/3=4x " "# (divide both sides by #4# to cancel it)

#8/3 = x #

BAM! You're done!

Dec 15, 2017

#x=2 and y=1#

Explanation:

This is the best form to have when solving a system of equations,!

They both show #y# in terms of #x#.

We know that #color(red)(y = y)#

So, the expressions that are both equal to #y# must be equal!

#color(white)(xxxx)color(red)(y = y)#

#color(red)(4x-7 = -7x+15)" "larr# solve for #x#

#4x +7x = 15+7#

#11x = 22#

#x = 2#

Substitute #2# for #x# in both equations to check that you get the same answer.

#y = 4x -7" "and" "y = -7x+15#

#y = 4(2) -7" "and" "y = -7(2)+15#

#y = 8-7" "and " "y = -14+15#

In both cases #y=1#