To determine which equation can replace [tex]\( 3x + 5y = 59 \)[/tex] in the original system while still producing the same solution, we will use the method of linear combination (also known as the elimination method).
The original system of equations is:
1) [tex]\( 3x + 5y = 59 \)[/tex]
2) [tex]\( 2x - y = -4 \)[/tex]
We aim to find another equation that can replace the first equation [tex]\( 3x + 5y = 59 \)[/tex] and still result in the same solution for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
Let's manipulate the second equation to see if we can form an equivalent equation.
Given:
[tex]\[ 2x - y = -4 \][/tex]
Multiply the entire equation by 5:
[tex]\[ 5 \cdot (2x - y) = 5 \cdot (-4) \][/tex]
This yields:
[tex]\[ 10x - 5y = -20 \][/tex]
Now we have:
[tex]\[ 10x - 5y = -20 \][/tex]
We need to check whether substituting [tex]\( 10x - 5y = -20 \)[/tex] in place of [tex]\( 3x + 5y = 59 \)[/tex] allows the system to retain the same solution for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
To summarize, the equation that can replace [tex]\( 3x + 5y = 59 \)[/tex] while still producing the same solution is:
[tex]\[ 10x - 5y = -20 \][/tex]
Thus, the correct equation is:
[tex]\[ \boxed{10x - 5y = -20} \][/tex]
