# i dont understand subtituting linear into quadratic

• Dec 10th 2006, 06:28 AM
kishan
i dont understand subtituting linear into quadratic
how would you substiture these two equations to find the values of x and y for these two questions.

1)y=x-5 2)3x-4y=25
y=2x^2-3x-3 and x^2+y^2=25

thank you
kishan
• Dec 10th 2006, 06:37 AM
Fredrik
Solve the linear equation for y (expressed as a function of x) and insert the result into the quadratic equation. The result is an equation you should be able to solve for x.

For example, in the second problem the equation you solve for x is x^2+((3x-25)/4))^2=25.
• Dec 10th 2006, 06:39 AM
kishan
plz
can u show it to me in steps because im confused
• Dec 10th 2006, 07:04 AM
Fredrik
Quote:

Originally Posted by kishan
can u show it to me in steps because im confused

OK, here's the first one:

The two equations can be combined to $x-5=2x^2-3x-3$. This is easy to simplify to $x^2-2x+1=0$, or if you prefer, $(x-1)^2=0$. The only solution is x=1. Use the first equation to find y:

$y=x-5=1-5=-4$

So the answer is x=1, y=-4.

The second problem:

Solve 3x-4y=25 for y. The result is y=(3x-25)/4. Insert this into the second equation:
$25=x^2+y^2=x^2+(\frac{3x-25}{4})^2$
$=x^2+\frac{1}{16}(9x^2+625-150x)$
$=\frac{25}{16}x^2-\frac{150}{16}x+\frac{625}{16}$

Multiply both sides with 16/25.

$16=x^2-6x+25$
$x^2-6x+9=0$
$(x-3)^2=0$
$x=3$
$y=(3*3-25)/4=-4$

So the answer is x=3, y=-4.