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Simultaneous equations. Help please?
I'm confused with this question because there is a y and a y^2.
Could you help please?
y=2x-3
y^2=8x-16
3 Answers
- Anonymous1 decade agoFavorite Answer
take the value of y in the first equation, that is 2x-3) and put it into the second equation in place of y. Now the second equation is all in x's and you can solve.
(2x-3)^2 = 8x - 16
4x^2 -12x + 9 = 8x - 16
4x^2 - 20x + 25 = 0
(2x-5)^2 = 0
2x-5 = 0
2x = 5
x = 5/2
Now take this x and put it back into the first equation and you can solve for y.
- 1 decade ago
(y=2x-3)^2
y^2=(2x-3)*(2x-3)
=4x^2-12x+9
must be equal to 8x-16 from eq. 2
4x^2-20x+25=0 factors to (2x-5)^2=0
The two equations are consistent only when x=5/2
