SOLUTION: Id be more than greatful if someone could help me with these two problems;
(I should just solve and check) but they are never equal when I solve them.
x-12=-8(x-3)
x^2-3x+5
Algebra ->
Exponents-negative-and-fractional
-> SOLUTION: Id be more than greatful if someone could help me with these two problems;
(I should just solve and check) but they are never equal when I solve them.
x-12=-8(x-3)
x^2-3x+5
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Question 179145: Id be more than greatful if someone could help me with these two problems;
(I should just solve and check) but they are never equal when I solve them.
x-12=-8(x-3)
x^2-3x+5= x(x-8)+20 Answer by monika_p(71) (Show Source):
You can put this solution on YOUR website! x-12=-8(x-3)
x-12=-8x+24 add 8x to both sides
x+8x-12=-8x+8x+24
9x-12=24 add 12
9x-12+12=24+12
9x=36 divide both sides by 9
x=4
Check:
4-12=-8(4-3)
-8=-8
-----------------------
x^2-3x+5= x(x-8)+20
x^2-3x+5=x^2-8x+20 subtract x^2 from both sides
x^2-x^2-3x+5=x^2-x^2-8x+20
-3x+5=-8x+20 add 8x to both sides
-3x+8x+5=-8x+8x+20
5x+5=20 subtract 5
5x+5-5=20-5
5x=15 divide by 5
x=3
Check:
3^2-3*3+5=3(3-8)+20
9-9+5=-15+20
5=5
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