Let's solve the given system of linear equations using the elimination method and determine the appropriate factor to multiply the second equation by so that the [tex]\( x \)[/tex]-terms will cancel out:
The system of equations is:
[tex]\[
\begin{array}{l}
-3x - 7y = -56 \\
-7x + 10y = 1
\end{array}
\][/tex]
Step 1: Multiply the first equation by a number that will make the coefficients of [tex]\( x \)[/tex] equal in absolute value.
The student decided to multiply the top equation by [tex]\( 7 \)[/tex]. Let's multiply the first equation by [tex]\( 7 \)[/tex]:
[tex]\[
7(-3x - 7y) = 7(-56)
\][/tex]
This simplifies to:
[tex]\[
-21x - 49y = -392
\][/tex]
Step 2: Identify the appropriate factor to multiply the second equation by so that the [tex]\( x \)[/tex]-terms will have the same or opposite coefficients.
To eliminate the [tex]\( x \)[/tex]-terms, we need the coefficient of [tex]\( x \)[/tex] in the second equation to be [tex]\( 21 \)[/tex] with the same sign as in the modified first equation if we want to subtract, or [tex]\( 21 \)[/tex] with opposite sign if we want to add.
We want the coefficients of [tex]\( x \)[/tex] to be perfectly opposite so that they cancel out when we add the equations.
We notice that the second equation has a term [tex]\( -7x \)[/tex]. To make this term [tex]\( 21x \)[/tex], we should multiply it by [tex]\( 3 \)[/tex]:
[tex]\[
3(-7x + 10y) = 3(1)
\][/tex]
This simplifies to:
[tex]\[
-21x + 30y = 3
\][/tex]
Step 3: Add the modified versions of the two equations to eliminate the [tex]\( x \)[/tex]-terms:
Add the resulting equations:
[tex]\[
\begin{array}{l}
(-21x - 49y) + (-21x + 30y) = -392 + 3 \\
-21x - 21x + (-49y + 30y) = -392 + 3 \\
-42x - 19y = -389
\end{array}
\][/tex]
Since we were able to align the coefficient of [tex]\( x \)[/tex] in both equations using the factors [tex]\( 7 \)[/tex] and [tex]\( 3 \)[/tex] respectively, the appropriate factor to multiply the second equation by is:
[tex]\[
\boxed{3}
\][/tex]
