The step is
Multiply 3 with eq(1) and multiply -4 with eq(3)
Then x coefficient becomes 12 and -12 respectively so you can cancel then out
Answer:
Multiply the top equation by -3 and the bottom equation by 2
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}4x-2y=7\\3x-3y=15 \end{cases}[/tex]
To solve the given system of equations by addition, make one of the variables in both equations sum to zero. To do this, the chosen variable must have the same coefficient, but it should be negative in one equation and positive in the other, so that when the two equations are added together, the variable is eliminated.
To eliminate the variable y:
Multiply the top equation by -3 to make the coefficient of the y variable 6:
[tex]\implies -3(4x-2y=7) \implies -12x+6y=-21[/tex]
Multiply the bottom equation by 2 to make the coefficient of the y variable -6:
[tex]\implies 2(3x-3y=15) \implies 6x-6y=30[/tex]
Add the two equations together to eliminate y:
[tex]\begin{array}{l r r}& -12x+6y= &-21\\+ & 6x-6y= & 30\\\cline{1-3}& -6x\phantom{))))))} = & 9\end{array}[/tex]
Solve for x:
[tex]\implies -6x=9[/tex]
[tex]\implies x=-\dfrac{9}{6}=-\dfrac{3}{2}[/tex]
Substitute the found value of x into one of the equations and solve for y:
[tex]\implies 3\left(-\dfrac{3}{2}\right)-3y=15[/tex]
[tex]\implies -\dfrac{9}{2}-3y=15[/tex]
[tex]\implies -3y=\dfrac{39}{2}[/tex]
[tex]\implies y=-\dfrac{39}{2 \cdot 3}[/tex]
[tex]\implies y=-\dfrac{13}{2}[/tex]
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