Q49. - Algebra 2

Question

Solve each system of equations by using either substitution or elimination

49. $\begin{array}{c}6x+y=15\\ x-4y=-10\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(x,y\right)=\left(2,3\right)$.

Step 1 – Apply elimination method to solve the system of equations.

Elimination method enables us to eliminate or get rid of one of the variable, so we can solve a more simplified equation.

Step 2 – Evaluate the system of equations

Given system of equations:

$\begin{array}{c}6x+y=15\\ x-4y=-10\end{array}$

Multiply the first equation, $6x+y=15$ by 4 and add the new resultant equation to the equation is $x-4y=-10$.

$\begin{array}{l}\underset{¯}{\begin{array}{c}6x+y=15\\ x-4y=-10\end{array}}\begin{array}{c}\text{multiply by 4}\to \\ \end{array}\underset{¯}{\begin{array}{c}24x+4y=60\\ x-4y=-10\end{array}}\\ 25x+0y=50\end{array}$

Solve the equation $25x=50$ for x:

 $\begin{array}{c}25x=50\\ \frac{25x}{25}=\frac{50}{25}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}divide both sides by 25}\\ x=2\end{array}$

Step 3 – Find the value of y

Substitute $x=2$ in $6x+y=15$ and solve for $y:$

 $\begin{array}{c}6x+y=15\\ 6\left(2\right)+y=15\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}substitute 2 for }x\\ 12+y=15\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}simplify}\\ y=3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}subtract 12 from both sides}\end{array}$

So, the solution of the given system is $\left(x,y\right)=\left(2,3\right)$.