How do solve the following linear system?:  5x+y=6 , 12x-2y=-16 ?

May 12, 2018

The solution to the linear system is $\left(- \frac{2}{11} , \frac{76}{11}\right)$.

The approximate solution is $\left(- 0.18 , 6.9\right)$.

Explanation:

Solve the linear system:

$\text{Equation 1} :$ $5 x + y = 6$

$\text{Equation 2} :$ $12 x - 2 y = - 16$

We can solve the system by elimination.

Multiply Equation 1 by $2$.

$2 \left(5 x + y = 6\right)$

$10 x + 2 y = 12$

Add Equation 1 and Equation 2.

$10 x + 2 y = \textcolor{w h i t e}{\ldots .} 12$
$12 x - 2 y = - 16$
$- - - - - - -$
$22 x \textcolor{w h i t e}{\ldots \ldots . .} = - 4$

Divide both sides by $22$.

$x = - \frac{4}{22}$

Simplify.

$x = - \frac{2}{11}$ or $\approx - 0.18$

Substitute the value for $x$ into Equation 1. Solve for $y$.

$5 x + y = 6$

$5 \left(- \frac{2}{11}\right) + y = 6$

Expand.

$- \frac{10}{11} + y = 6$

Add $\frac{10}{11}$ to both sides.

$y = 6 + \frac{10}{11}$

Multiply $6$ by $\frac{11}{11}$ to get an equivalent fraction with $11$ as the denominator.

$y = 6 \times \frac{11}{11} + \frac{10}{11}$

$y = \frac{66}{11} + \frac{10}{11}$

$y = \frac{76}{11}$ or $\approx 6.9$

The solution to the linear system is $\left(- \frac{2}{11} , \frac{76}{11}\right)$.

The approximate solution is $\left(- 0.18 , 6.9\right)$.

graph{(5x+y-6)(12x-2y+16)=0 [-7.52, 6.53, 1.603, 8.627]}

May 12, 2018

$x = - \frac{2}{11}$
$y = \frac{76}{11}$

Explanation:

Given -

$5 x + y = 6$ --------- (1)
$12 x - 2 y = - 16$ ------(2)

$5 x + y = 6$ --------- (1) $\times 2$
$12 x - 2 y = - 16$ ------(2)

$10 x + 2 y = 12$--------(3)
$12 x - 2 y = - 16$ -------(2) -- $\left(3\right) + \left(2\right)$
$22 x = - 4$
$x = - \frac{4}{22} = - \frac{2}{11}$

$x = - \frac{2}{11}$

Plug in $x = - \frac{2}{11}$ in equation (1)

$5 \left(- \frac{2}{11}\right) + y = 6$
$- \frac{10}{11} + y = 6$

$y = 6 + \frac{10}{11} = \frac{66 + 10}{11} = \frac{76}{11}$

$y = \frac{76}{11}$

May 12, 2018

$y = \frac{76}{11}$

$x = - \frac{2}{11}$

Explanation:

Many ways, but my favorite is the elimination method, and fortunately, it works in this situation!

Let's make the equations look neater (put in $y = m x + b$ form)

• Equation 1: $\text{ } 5 x + y = 6 \mathmr{and} y = - 5 x + 6$
• Equation 2: $\text{ } 12 x - 2 y = - 16 \mathmr{and} y = 6 x + 8$

The elimination method allows us to cancel out one of the variables, making it an easy algebra equation to solve for the remaining variable. You'll see.

Let's eliminate the $7$ variable. In order to do that, we need to simply multiply one of the equations by negative one, let's do it to equation one.

• Equation 1: $\text{ } - y = + 5 x - 6$
• Equation 2: $\text{ } y = 6 x + 8$

Now, we add the equations together, getting one combined equation:

$0 = 11 x + 2$

$x = - \frac{2}{11}$

Now, we plug in the $x$ value into one of the ORIGINAL equations to solve for $y$.

Equaiton 1:

$y = - 5 \left(- \frac{2}{11}\right) + 6$

$y = \frac{76}{11}$