How do you solve the following system?: #61x -31y =-33 , -3x +7y = 8#

We have:

#{(61x-31y=-33" "" "" "" "" "mathbf(eq. 1)),(-3x+7y=8" "" "" "" "" "" "color(white)(sl)mathbf(eq. 2)):}#

We want to either cancel out the #x# terms or the #y# terms. We can cancel the #x# terms by multiplying #mathbf(eq. 1)# by #3# and #mathbf(eq. 2)# by #61#, and then adding the two.

#{:(color(white)"-.-"183x-93y=-99" "" "" "" "color(white)(sl)mathbf(eq. 1)xx3),(ul(-183x+427y=488" "+)" "" "color(white)(ss)mathbf(eq. 2)xx61),(color(white)("-------------")334y=389" "" "" "" "" "mathbf(eq. 3)):}#

From #mathbf(eq. 3)#, we can solve for #y# to see that

#color(blue)(y=389/334#

We can now plug this value of #y# into either of the original two equations. I will choose #mathbf(eq. 2)# since the numbers will be smaller.

#-3x+7color(blue)y=8" "=>" "-3x+7(color(blue)(389/334))=8#

Solving this equation, we multiply #7# and #389//334#:

#-3x+2723/334=8#

#-3x=8-2723/334#

Find a common denominator.

#-3x=2672/334-2723/334#

#-3x=-51/334#

#x=-51/334(-1/3)#

The negatives will cancel, so #x# will be positive, and note that #51=3xx17#, so the final value of #x# is:

#color(red)(x=17/334#

Written as the ordered pair #(x,y)#, our final answer is

#color(green)(|barul(color(white)(int^int)(x,y)=(17/334,389/334)color(white)(int^int)|))#