The system of equations can be solved using linear combination to eliminate one of the variables.

Which equation can replace [tex]3x + 5y = 59[/tex] in the original system and still produce the same solution?

A. [tex]2x - y = -4[/tex]
B. [tex]10x - 5y = -20[/tex]
C. [tex]7x = 39[/tex]
D. [tex]13x = 39[/tex]



Answer :

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]