How do you solve the simultaneous equations 3x + 5y = 113x+5y=11 and 2x + y = 32x+y=3?

2 Answers
Gió
Jul 22, 2015

I found:
x=4/7x=47
y=13/7y=137

Explanation:

From the second equation you can isolate yy as:
y=3-2xy=3−2x
substitute nto the first to find xx:
3x+5(color(red)(3-2x))=113x+5(3−2x)=11
3x+15-10x=113x+15−10x=11
-7x=-4−7x=−4
x=4/7x=47
substitute back into the second equation:
y=3-2(4/7)=(21-8)/7=13/7y=3−2(47)=21−87=137

Alan P.
Jul 22, 2015

(x,y) = (4/7, 13/7)(x,y)=(47,137)

Explanation:

[1]color(white)("XXXX")XXXX3x+5y = 113x+5y=11
[2]color(white)("XXXX")XXXX2x+y = 32x+y=3

subtract 2x2x from both sides of [2] to isolate yy on the left side
[3]color(white)("XXXX")XXXXy = 3-2xy=3−2x

substitute (3-2x)(3−2x) for yy in [1]
[4]color(white)("XXXX")XXXX3x + 5(3-2x) = 113x+5(3−2x)=11

simplify
[5]color(white)("XXXX")XXXX-7x + 15 = 11−7x+15=11

[6]color(white)("XXXX")XXXXx = 4/7x=47

substituting 4/747 for xx in [2]
[7]color(white)("XXXX")XXXX2(4/7) + y = 32(47)+y=3

[8]color(white)("XXXX")XXXXy = 13/7y=137

Related questions
See all questions in Systems Using Substitution
Impact of this question
8241 views around the world
You can reuse this answer
Creative Commons License