Answer :
To use the elimination method, the coefficient of one variable in one equation must be the opposite number of the coefficient of the same variable in the second equation.
[tex]4x-3y=8 \\ 5x-2y=-11[/tex]
Here you can multiply the first equation by 2, and the second equation by -3.
[tex]4x-3y=8 \ \ |\cdot 2 \\ 5x-2y=-11 \ \ |\cdot (-3) \\ \\ 8x-6y=16 \\ -15x+6y=33[/tex]
Now you just add the equations by sides and solve for one variable.
[tex]8x-6y=16 \\ \underline{-15x+6y=33 } \\ 8x-6y-15x+6y=16+33 \\ 8x-15x=16+33 \\ -7x=49 \\ x=-7[/tex]
Now you solve for the other variable by substituting -7 for x in one of the equations.
[tex]4x-3y=8 \\ 4 \cdot (-7)-3y=8 \\ -28-3y=8 \\ -3y=8+28 \\ -3y=36 \\ y=-12[/tex]
You could choose x at the beginning as well. Then you'd have:
[tex]4x-3y=8 \ \ |\cdot (-5)\\ 5x-2y=-11 \ \ |\cdot 4 \\ \\ -20x+15y=-40 \\ \underline{20x-8y=-44 \ \ \ \ \ \ } \\ -20x+15y+20x-8y=-40-44 \\ 15y-8y=-40-44 \\ 7y=-84 \\ y=-12 \\ \\ 4x-3y=8 \\ 4x-3 \cdot (-12)=8 \\ 4x+36=8 \\ 4x=8-36 \\ 4x=-28 \\ x=-7[/tex]
So the answer is:
[tex]x=-7 \\ y=-12[/tex]
[tex]4x-3y=8 \\ 5x-2y=-11[/tex]
Here you can multiply the first equation by 2, and the second equation by -3.
[tex]4x-3y=8 \ \ |\cdot 2 \\ 5x-2y=-11 \ \ |\cdot (-3) \\ \\ 8x-6y=16 \\ -15x+6y=33[/tex]
Now you just add the equations by sides and solve for one variable.
[tex]8x-6y=16 \\ \underline{-15x+6y=33 } \\ 8x-6y-15x+6y=16+33 \\ 8x-15x=16+33 \\ -7x=49 \\ x=-7[/tex]
Now you solve for the other variable by substituting -7 for x in one of the equations.
[tex]4x-3y=8 \\ 4 \cdot (-7)-3y=8 \\ -28-3y=8 \\ -3y=8+28 \\ -3y=36 \\ y=-12[/tex]
You could choose x at the beginning as well. Then you'd have:
[tex]4x-3y=8 \ \ |\cdot (-5)\\ 5x-2y=-11 \ \ |\cdot 4 \\ \\ -20x+15y=-40 \\ \underline{20x-8y=-44 \ \ \ \ \ \ } \\ -20x+15y+20x-8y=-40-44 \\ 15y-8y=-40-44 \\ 7y=-84 \\ y=-12 \\ \\ 4x-3y=8 \\ 4x-3 \cdot (-12)=8 \\ 4x+36=8 \\ 4x=8-36 \\ 4x=-28 \\ x=-7[/tex]
So the answer is:
[tex]x=-7 \\ y=-12[/tex]
