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

##### 3 Answers
Aug 3, 2018

${x}^{2} + {y}^{2} = 4$

#### Explanation:

Here,

$3 \left(4 - 2 x\right) = 0$

Dividing both sides by $3$ ,we get

$4 - 2 x = 0$

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

Adding $4$ both sides ,

$2 x - 4 + 4 = 0 + 4$

$\therefore 2 x = 4$

Dividing both sides by $2$ ,we get

color(red)(x=2

Again,

$2 y + 4 = 4$

Adding $\left(- 4\right)$ both sides

$2 y + 4 + \left(- 4\right) = 4 + \left(- 4\right)$

$\therefore 2 y = 0$

Dividing both sides by $2$ ,we get

color(red)(y=0

So ,

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

$\therefore {x}^{2} + {y}^{2} = 4$

Aug 3, 2018

The value of ${x}^{2} + {y}^{2}$ is $4$.

#### Explanation:

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

$3 \left(4 - 2 x\right) = 0$

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

Subtract $\textcolor{b l u e}{12}$ from both sides:
$12 - 6 x \quad \textcolor{b l u e}{- \quad 12} = 0 \quad \textcolor{b l u e}{- \quad 12}$

$- 6 x = - 12$

Divide both sides by color(blue(-6):
$\frac{- 6 x}{\textcolor{b l u e}{- 6}} = \frac{- 12}{\textcolor{b l u e}{- 6}}$

$\textcolor{red}{x = 2}$

$2 y + 4 = 4$

Subtract $\textcolor{b l u e}{4}$ from both sides:
$2 y + 4 \quad \textcolor{b l u e}{- \quad 4} = 4 \quad \textcolor{b l u e}{- \quad 4}$

$2 y = 0$

Divide both sides by $\textcolor{b l u e}{2}$:
$\frac{2 y}{\textcolor{b l u e}{2}} = \frac{0}{\textcolor{b l u e}{2}}$

$\textcolor{red}{y = 0}$

We want the value of ${x}^{2} + {y}^{2}$, so:
${2}^{2} + {0}^{2} = 4 + 0 = \textcolor{red}{4}$

Hope this helps!

Aug 4, 2018

${x}^{2} + {y}^{2} = 4$

#### 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:

$3 \left(4 - 2 x\right) = 0$

We can divide both sides by $3$ to get

$- 2 x + 4 = 0 \implies - 2 x = - 4 \implies \textcolor{b l u e}{x = 2}$

We also have

$2 y + 4 = 4$

We can subtract $4$ from both sides to get

$2 y = 0 \implies \textcolor{red}{y = 0}$

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

$\textcolor{b l u e}{{2}^{2}} + \textcolor{red}{{0}^{2}} = 4 + 0 = 4$

Hope this helps!