Q50. - Algebra 2

Question

Solve each system of equations by using either substitution or elimination 50 $\begin{array}{c}3x+8y=23\\ 5x-y=24\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is  .$(x,y)=(5,1)$

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:$\begin{array}{c}3x+8y=23\\ 5x-y=24\end{array}$

Multiply the second equation, $5x-y=24$ by 8 and add the new resultant equation to the equation is $3x+8y=23$

$\begin{array}{l}\underset{\_}{\begin{array}{c}3x+8y=23\\ 5x-y=24\end{array}}\begin{array}{c}\\ \text{multiply by 8}\to \end{array}\underset{\_}{\begin{array}{c}3x+8y=23\\ 40x-8y=192\end{array}}\\ 43x+0y=215\end{array}$

Solve the equation $43x=215$ for $x$$\begin{array}{c}43x=215\\ \frac{43x}{43}=\frac{215}{43}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divide both sides by 43}\\ x=5\end{array}$:

Step 3 – Find the value of y

Substitute $x=5$ in $5x-y=24$ and solve for $y:$

$\begin{array}{c}5x-y=24\\ 5\left(5\right)-y=24\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute 5 for }x\\ 25-y=24\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ -y=-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}subtract 25 from both sides}\\ y=1\text{ divide both sides by }-1\end{array}$

So, the solution of the given system is  .$(x,y)=(5,1)$