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3x - 4y = 10Add a Com6x + y = 38andMediac-9% +12y = -30Save9x - 3y = 48These systems are said to be equivalent. Both of the equations in the secondsystem came from the first system somehow,Two questions: How was the first equation is the second system formed fromthe first system? And how was the second equation in the second systemformed from the first system?

Answer :

[tex]\begin{gathered} First\text{ System of Equations} \\ 3x\text{ - 4y = 10 }\ldots\ldots Equation\text{ 1}.1 \\ 6x\text{ + y = 3}8\ldots..Equation\text{ 1.2} \end{gathered}[/tex][tex]\begin{gathered} \text{Second System of Equations} \\ -9x\text{ + 12y = -30 }\ldots\ldots Equation\text{ 2.1} \\ 9x\text{ - 3y = 48 }\ldots\ldots Equation\text{ 2.2} \end{gathered}[/tex]

First solution

[tex]\begin{gathered} \text{The first equation in the second system (Equation 2.1),} \\ \text{was formed by multiplying (Equation 1.1) by -3} \end{gathered}[/tex][tex]\begin{gathered} \text{proof:} \\ -3\text{ (3x - 4y= 10) = -9x + 12y = -30} \end{gathered}[/tex]

Second solution

The second equation in the second system (Equation 2.2) was formed by adding both equations in the first system.

That is;

(Equation 2.2) = (Equation 1.1) + (Equation 1.2)

[tex]\begin{gathered} \text{proof:} \\ 3x\text{ -4y = 10 } \\ +\text{ 6x + y = 38} \\ (3x+6x)\text{ + (-4y+y) = 10+38} \\ 9x\text{ -3y = 48} \end{gathered}[/tex]

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