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To determine which equation can replace [tex]\( 3x + 5y = 59 \)[/tex] in the original system while still producing the same solution, consider the original system of linear equations:
1. [tex]\( 3x + 5y = 59 \)[/tex]
2. [tex]\( 2x - y = -4 \)[/tex]
First, let's manipulate the second equation to align it with the first equation in a way that helps us maintain the solutions' consistency.
Multiply the second equation by 5:
[tex]\[ 5(2x - y) = 5(-4) \][/tex]
Expanding this, we get:
[tex]\[ 10x - 5y = -20 \][/tex]
Now, observe that the equation [tex]\( 10x - 5y = -20 \)[/tex] can serve as a replacement for the original equation [tex]\( 3x + 5y = 59 \)[/tex] while maintaining the solutions' consistency. This new system of equations will now be:
1. [tex]\( 10x - 5y = -20 \)[/tex]
2. [tex]\( 2x - y = -4 \)[/tex]
These two equations will yield the same solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] as the original system.
Therefore, the equivalent equation that can replace [tex]\( 3x + 5y = 59 \)[/tex] is [tex]\( 10x - 5y = -20 \)[/tex]. Hence, the correct option is:
[tex]\[ 10x - 5y = -20 \][/tex]
1. [tex]\( 3x + 5y = 59 \)[/tex]
2. [tex]\( 2x - y = -4 \)[/tex]
First, let's manipulate the second equation to align it with the first equation in a way that helps us maintain the solutions' consistency.
Multiply the second equation by 5:
[tex]\[ 5(2x - y) = 5(-4) \][/tex]
Expanding this, we get:
[tex]\[ 10x - 5y = -20 \][/tex]
Now, observe that the equation [tex]\( 10x - 5y = -20 \)[/tex] can serve as a replacement for the original equation [tex]\( 3x + 5y = 59 \)[/tex] while maintaining the solutions' consistency. This new system of equations will now be:
1. [tex]\( 10x - 5y = -20 \)[/tex]
2. [tex]\( 2x - y = -4 \)[/tex]
These two equations will yield the same solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] as the original system.
Therefore, the equivalent equation that can replace [tex]\( 3x + 5y = 59 \)[/tex] is [tex]\( 10x - 5y = -20 \)[/tex]. Hence, the correct option is:
[tex]\[ 10x - 5y = -20 \][/tex]
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