Question

solve the system of equations $3x - 8y = 36$ and $2x + y = 5$ by combining the equations.
\\(\square\\) $(3x\\ \\ -8y\\ =36)$
\\(\square\\) $(2x\\ \\ +y\\ =\\ 5)$
\\(\

$$\begin{array}{r}3x\\ -8y\\ =\\ 36\\\\2x\\ +y\\ =\\ 5\\\\\\hline\\square\\ x+\\ \\square\\ y=\\ \\square\\end{array}$$

$ try
answer attempt 1 out of 5
you must answer all questions above in order to submit.

Explanation:

Step1: Multiply second equation by 8

Multiply each term in $2x + y = 5$ by 8 to eliminate $y$:
$$8(2x + y) = 8\times5 \implies 16x + 8y = 40$$

Step2: Add to first equation

Add the modified second equation to $3x - 8y = 36$:
$$(3x - 8y) + (16x + 8y) = 36 + 40$$
Simplify:
$$19x + 0y = 76$$

Step3: Solve for $x$

Isolate $x$ by dividing both sides by 19:
$$x = \frac{76}{19} = 4$$

Step4: Substitute $x=4$ to find $y$

Plug $x=4$ into $2x + y = 5$:
$$2(4) + y = 5 \implies 8 + y = 5$$
Solve for $y$:
$$y = 5 - 8 = -3$$

Answer:

The multipliers for the equations are $\boldsymbol{1}$ (for $3x -8y=36$) and $\boldsymbol{8}$ (for $2x + y=5$).
After combining: $19x + 0y = 76$
The solution to the system is $x=4$, $y=-3$