Question

If `3(4-2x) = 0` and `2y+4=4`, then what is the value of `x^2+y^2`?

Answer 1

`x^2+y^2=4`

Explanation:

Here,

`3(4-2x)=0`

Dividing both sides by `3` ,we get

`4-2x=0`

`i.e. 2x-4=0`

Adding `4` both sides ,

`2x-4+4=0+4`

`:.2x=4`

Dividing both sides by `2` ,we get

`color(red)(x=2`

Again,

`2y+4=4`

Adding `(-4)` both sides

`2y+4+(-4)=4+(-4)`

`:.2y=0`

Dividing both sides by `2` ,we get

`color(red)(y=0`

So ,

`x^2+y^2=color(red)((2)^2+(0)^2`

`:. x^2+y^2=4`

Answer 2

The value of `x^2 + y^2` is `4`.

Explanation:

Simplify both equations and solve for `x` and `y`:

`3(4-2x) = 0`

Distribute the right hand side:
`12 - 6x = 0`

Subtract `color(blue)12` from both sides:
`12 - 6x quadcolor(blue)(-quad12) = 0 quadcolor(blue)(-quad12)`

`-6x = -12`

Divide both sides by `color(blue(-6)`:
`(-6x)/color(blue)(-6) = (-12)/color(blue)(-6)`

`color(red)(x = 2)`

`2y + 4 = 4`

Subtract `color(blue)4` from both sides:
`2y + 4 quadcolor(blue)(-quad4) = 4 quadcolor(blue)(-quad4)`

`2y = 0`

Divide both sides by `color(blue)2`:
`(2y)/color(blue)2 = 0/color(blue)2`

`color(red)(y = 0)`

We want the value of `x^2 + y^2`, so:
`2^2 + 0^2 = 4 + 0 = color(red)(4)`

Hope this helps!

Answer 3

`x^2+y^2=4`

Explanation:

We can solve for `x` and `y`, and then square them to find the value of `x^2+y^2`.

We have the following:

`3(4-2x)=0`

We can divide both sides by `3` to get

`-2x+4=0=>-2x=-4=>color(blue)(x=2)`

We also have

`2y+4=4`

We can subtract `4` from both sides to get

`2y=0=>color(red)(y=0)`

Now, let's plug these values into `x^2+y^2` to get

`color(blue)(2^2)+color(red)(0^2)=4+0=4`

Hope this helps!