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Solving systems of linear equations using substitution.... y=x-3; x+y=5
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use substitution to isolate one variable (already done here) and use it to solve the other x+(x-3) = 5 x + x = 8 2x = 8 x = 4 use that to solve first half y = 4 - 3 y = 1
2x - 3 = 5 x = 4 sub in isr equn, y = 4-3 = 1 x=4, y=1
How did you come up with 8?
add three to both sides
you need to solve for x so you need x by itself
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you could do this the other way around and solve for y first too, I just used x first
Thank you, that is very helpful
yw
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