How do you solve -x-3y= 15−x−3y=15 and 2x+7y= -362x+7y=−36 using substitution?

2 Answers
Professor Sam
May 22, 2017

x = 3x=3

Explanation:

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

3y = 15 + -x3y=15+−x

-x = 15+3y−x=15+3y

(-x)/-1 = (15+3y)/-1−x−1=15+3y−1

x = (15+3y)/-1x=15+3y−1

(15+3y)/-1xx(-1 )/-1 =( -15 + -3y)/115+3y−1×−1−1=−15+−3y1

2x + 7y = -362x+7y=−36

Substitute -15 + -3y−15+−3y for x

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

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

-30+ -6y +7y = -36−30+−6y+7y=−36

-30 + y = -36−30+y=−36

y = -6y=−6

Plug in the y's value

-x - 3 * -6 = 15−x−3⋅−6=15

-x - (-18) = 15−x−(−18)=15

-x + 18 = 15−x+18=15

-x = 15-18−x=15−18

-x = -3−x=−3

-(-x) = -(-3)−(−x)=−(−3)

x = 3x=3

But an easier approach would be using elimination method.

Jim G.
May 22, 2017

3,-6)3,−6)

Explanation:

-color(red)(x)-3y=15to(1)−x−3y=15→(1)

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

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

rArrcolor(red)(x)=-3y-15to(3)⇒x=−3y−15→(3)

"substitute this value into " (2)substitute this value into (2)

rArr2(-3y-15)+7y=-36⇒2(−3y−15)+7y=−36

rArr-6y-30+7y=-36larr" distributing"⇒−6y−30+7y=−36← distributing

rArry-30=-36larr" simplifying left side"⇒y−30=−36← simplifying left side

"add 30 to both sides"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)

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