# How do you solve 4x – y = 5 and x + y = 10 and which method do you use?

May 29, 2018

$x = 3$ and $y = 7$

#### Explanation:

First take the second equation...
$x + y = 10$
$x = 10 - y$ ...$\left(e q . 1\right)$

Now substitute $\left(e q . 1\right)$ in the given equation $4 x - y = 5$.

$4 \left(10 - y\right) - y = 5$

$40 - 4 y - y = 5$

$40 - 5 y = 5$

$40 - 5 = 5 y$

$35 = 5 y$

$y = 7$

Substituting this value of $y$ in $\left(e q . 1\right)$, we get $x = 3$.

Therefore, $x = 3$ and $y = 7$.

May 29, 2018

$\left(3 , 7\right)$

#### Explanation:

$4 x - y = 5 - - \left(1\right)$

$x + y = 10 - - \left(2\right)$

solve by elimination

$\left(1\right) + \left(2\right)$

$5 x = 15$

$\implies x = 3$

substitute into $\left(2\right)$

$3 + y = 10$

$\implies y = 7$

check in $\left(1\right)$

$4 \times 3 - 7 = 12 - 7 = 5$

$= R H S$(1)#

$\therefore$consistent

$\left(3 , 7\right)$

May 29, 2018

See explanation.

The system is:

## $\left\{\begin{matrix}4 x - y = 5 \\ x + y = 10\end{matrix}\right.$

It can be solved using any of 3 methods:

• Using substitution:

From the second equation we can calculate that: $y = 10 - x$

If we put this in the first equation we get:

## $5 x = 15 \implies x = 3$

Now we can calculate that $y = 10 - 3 = 7$
So the solution is

$\left\{\begin{matrix}x = 3 \\ y = 7\end{matrix}\right.$

In the initial system the coefficients of $y$ are opposite numbers $- 1$ and $1$, so if we add both equations we get an equation with $x$ variable only:

$5 x = 15$
$x = 3$

Now we can calculate the remaining variable $y$ by substitution:

$3 + y = 10 \implies y = 7$

• Graphically

Both equations represent linear functions, so we can solve the system by graphing the lines and seeing if they intersect:

graph{(y-4x+5)(x+y-10)((x-3)^2+(y-7)^2-0.05)=0 [-10, 10, -8, 8]}

As we can see the lines intersect at $\left(3 , 7\right)$, so the solution is:

$\left\{\begin{matrix}x = 3 \\ y = 7\end{matrix}\right.$

The choice of method depends on the system of equations. Here the easiest (for me) is the second method but others may prefer different ones.