How do you solve #-x-3y= 15# and #2x+7y= -36# using substitution?

2 Answers
May 22, 2017

Answer:

#x = 3#

Explanation:

#-x - 3y = 15" "("Solve for "x)#

#3y = 15 + -x#

#-x = 15+3y#

#(-x)/-1 = (15+3y)/-1#

#x = (15+3y)/-1#

#(15+3y)/-1xx(-1 )/-1 =( -15 + -3y)/1#

#2x + 7y = -36#

Substitute #-15 + -3y# for x

#2*(-15+ -3y)+7y =-36"(Distribute)"#

#(2*- 15) + (2 *3y) +7y = -36#

# -30+ -6y +7y = -36#

#-30 + y = -36#

#y = -6#

Plug in the y's value

#-x - 3 * -6 = 15#

#-x - (-18) = 15#

#-x + 18 = 15#

#-x = 15-18#

#-x = -3#

#-(-x) = -(-3)#

#x = 3#

But an easier approach would be using elimination method.

May 22, 2017

Answer:

#3,-6)#

Explanation:

#-color(red)(x)-3y=15to(1)#

#2color(red)(x)+7y=-36to(2)#

#"from " (1)" we can make x the subject"#

#rArrcolor(red)(x)=-3y-15to(3)#

#"substitute this value into " (2)#

#rArr2(-3y-15)+7y=-36#

#rArr-6y-30+7y=-36larr" distributing"#

#rArry-30=-36larr" simplifying left side"#

#"add 30 to both sides"#

#ycancel(-30)cancel(+30)=-36+30#

#rArry=-6#

#"substitute into " (3)" and evaluate for x"#

#x=(-3xx-6)-15=18-15=3#

#rArr"point of intersection "=(3,-6)#