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# Solve the system: `x^2-4y^2=16`  ```x-6y=-4`

kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 15, 2013 at 12:52 AM via web

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Solve the system:

`x^2-4y^2=16`

```x-6y=-4`

Tagged with algebra 2, math

tjbrewer | Elementary School Teacher | (Level 2) Associate Educator

Posted July 15, 2013 at 1:39 AM (Answer #1)

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To solve the equation, we solve the second equation for x.  `x-6y=-4->x=6y-4`

Now we substitute our value for x, into the first equation.

`x^2-4y^2=16->(6y-4)^2-4y^2=16`

We simplify and we get

`36y^2-48y+16-4y^2=16`

Combine like terms, and solve for 0

`32y^2-48y=0`

The most obvious answer is that y=0, and that is one solution.  The other value is found by rewriting the equation:

`32y^2=48y`

we divide each side by y and get

`32y=48`

We solve for y by dividing 48 by 32 and  `y=48/32=3/2=1 1/2`

now we substitute these y values back into one of our original equations to get x values.

`x-6xx3/2=-4->x-9=-4->x=5`

`x-6xx0=-4->x-0=-4->x=-4`

So our answers are ``and `x=-4, 5`` `

Or in ordered pairs (-4,0) and (5, 1.5)

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