# How do solve the following linear system?:  x= y - 11 , y = 4x + 4 ?

Jun 29, 2016

Soln. is $\left(x , y\right) = \left(\frac{7}{3} , \frac{40}{3}\right) .$

#### Explanation:

Submit the value of $y ,$ taken from ${2}^{n d}$ eqn., in the ${1}^{s t}$ eqn. to get,

$x = 4 x + 4 - 11 \Rightarrow x - 4 x = 4 - 11 \Rightarrow - 3 x = - 7 \Rightarrow x = - \frac{7}{-} 3 = \frac{7}{3.}$

Hence, by ${2}^{n d}$ eqn., $y = 4 \left(\frac{7}{3}\right) + 4 = \frac{28}{3} + 4 = \frac{40}{3.}$

Soln. is $\left(x , y\right) = \left(\frac{7}{3} , \frac{40}{3}\right) .$

Jun 29, 2016

$x = \frac{7}{3} = 2 \frac{1}{3} \mathmr{and} y = \frac{40}{3} = 13 \frac{1}{3}$

#### Explanation:

There are several methods you can use, but there is a single $y$ term in each equation, so equating them is probably the easiest method.

Change the first equation to: $y = x + 11$

Now, the y in each equation is the same, so;

if: $\text{ } y = y$
then:$\text{ } 4 x + 4 = x + 11$

$\text{ } 4 x - x = 11 - 4$
$\text{ } 3 x = 7$

$x = \frac{7}{3} = 2 \frac{1}{3}$

To find the value of $y$, substitute $x = \frac{7}{3}$ into either of the equations above ..... or both, to check that they give the same answer.

$y = 4 \times \frac{7}{3} + 4 = \frac{40}{3} \text{ and } y = \frac{7}{3} + 11 = \frac{40}{3}$