Question

$x+y+3z=600$
$x+y+z=400$
In the system of equations above, what is the value of $x+y$ ?

Answer 1

To find the value of $x$ $+$ $y$

Let us assume $(x$ $+$ $y)$ $=$ $t$

Given, $x$ $+$ $y$ $+$ $3z$ $=$ $600$

Substituting $(x$ $+$ $y)$ $=$ $t$, we get

$t$ $+$ $3z$ $=$ $600$

$x$ $+$ $y$ $+$ $z$ $=$ $400$

Substituting $(x$ $+$ $y)$ $=$ $t$, we get

$t$ $+$ $z$ $=$ $400$

Rearranging the terms, we get

$z$ $=$ $400$ $-$ $t$

From the above,

$t$ $+$ $(3$ $\times$ $z)$ $=$ $600$

Substituting $z$ $=$ $400$ $-$ $t$, we get

$t$ $+$ $(3$ $\times$ $(400$ $-$ $t\left)\right)$ $=$ $600$

$\Rightarrow$ $t+$ $1200$ $-3t$ $=$ $600$

Rearranging the terms, we get

$2$ $\times$ $t$ $=$ $600$

$\Rightarrow t$ $=$ $\frac{600}{2}$

$\Rightarrow t$ $=$ $300$

$\Rightarrow t$ $=$ $(x$ $+$ $y)$ $=$ $300$

Therefore, the value of ${}^{\prime }(x$ $+$ $y{)}^{\prime }$ is ${}^{\prime }{300}^{\prime }$.