Solve the following systems using the elimination method.
\begin{cases} 2x-y=4 \cr \cr 4x+2y=3 \end{cases}
Multiply the equations by a constant
Multiplying the top equation 2x-y = 4 by 2 forces the coefficient of the y's to become the opposites of each other.
The equations become:
\begin{cases} 4x-2y=8 \cr \cr 4x+2y=3 \end{cases}
Add the equations
Adding the two equations gives:
8x=11
Divide by 8:
\dfrac{8x}{8}=\dfrac{11}{8}
x=\dfrac{11}{8}
Substitute to solve for the other variable
Substituting x=\dfrac{11}{8} into 4x + 2y = 3 gives:
\dfrac{11}{2}+2y=3
Subtract \dfrac{11}{2} from both sides:
\dfrac {11} {2} + 2y- \dfrac {11} {2} = 3- \dfrac {11} {2}
2y=\dfrac{-5}{2}
Multiply both sides by \dfrac{1}{2} :
y=\dfrac{-5}{4}
The solution of this system is \left(\dfrac{11}{8},\dfrac{-5}{4}\right).
\begin{cases} 3x-2y=7 \cr \cr 5x+2y=9 \end{cases}
Add the equations
Adding the two equations gives:
8x=16
Divide both sides by 8:
x=2
Substitute to solve for the other variable
Substituting x=2 into 3x-2y=7 gives:
3\left(2\right)-2y=7
6-2y=7
Subtract 6 from both sides:
-2y=1
Divide both sides by -2:
y=\dfrac{-1}{2}
The solution of this system is \left(2,\dfrac{-1}{2}\right).
\begin{cases} 7y-4x=4 \cr \cr 4x+3y=6 \end{cases}
Line up the x's and y's
\begin{cases} 7y-4x=4 \cr \cr 4x+3y=6 \end{cases}
Becomes:
\begin{cases} -4x+7y=4 \cr \cr 4x+3y=6 \end{cases}
Add the equations
Adding the two equations gives:
10y=10
Divide both sides by 10:
y=1
Substitute to solve for the other variable
Substituting y=1 into 4x+3y=6 gives:
4x+3\left(1\right)=6
4x+3=6
Subtracting 3 from both sides gives:
4x=3
Dividing both sides by 4 gives:
x=\dfrac{3}{4}
The solution of this system is \left(\dfrac{3}{4},1\right).
\begin{cases} 5x-3y=8 \cr \cr 5x+2y=3 \end{cases}
Multiply the equations by a constant
Multiplying the top equation 5x-3y=8 by -1 forces the coefficient of the y's to become the opposites of each other.
The equations become:
\begin{cases} -5x+3y=-8 \cr \cr 5x+2y=3 \end{cases}
Add the equations
Adding the two equations gives:
5y=-5
Dividing both sides by 5 gives:
y=-1
Substitute to solve for the other variable
Substitute y=-1 into 5x+2y=3 :
5x+2\left(-1\right)=3
5x-2=3
5x=5
Divide by 5:
x=1
The solution of this system is \left(1,-1\right).
\begin{cases} y-4x=10 \cr \cr 2x+2y=0 \end{cases}
Line up the x's and y's
\begin{cases} y-4x=10 \cr \cr 2x+2y=0 \end{cases}
Becomes:
\begin{cases}-4x+y=10 \cr \cr 2x+2y=0 \end{cases}
Multiply the equations by a constant
Multiplying the equation 2x+2y=0 by 2 forces the coefficients to be the opposites of each other:
\begin{cases}-4x+y=10 \cr \cr 4x+4y=0 \end{cases}
Add the equations
Adding the two equations gives:
5y=10
Divide both sides by 5:
y=2
Substitute to solve for the other variable
Substituting y=2 into 2x+2y=0 gives:
2x+2\left(2\right)=0
2x+4=0
Subtract 4 from both sides:
2x=-4
Divide both sides by 2:
x=-2
The solution of this system is \left(-2{,}2\right).
\begin{cases} 9x-2y=17 \cr \cr 5x+2y=12 \end{cases}
Add the equations
Adding the two equations gives:
14x=29
Dividing both sides by 14 gives:
x=\dfrac{29}{14}
Substitute to solve for the other variable
Substituting x=\dfrac{29}{14} into 5x+2y=12 gives:
5\left(\dfrac{29}{14}\right)+2y=12
\dfrac{145}{14}+2y=12
Subtracting \dfrac{145}{14} to both sides gives:
2y=\dfrac{23}{14}
Multiply both sides by \dfrac{1}{2} :
y=\dfrac{23}{28}
The solution of this system is \left(\dfrac{29}{14},\dfrac{23}{28}\right).
\begin{cases} 3x-4y=12 \cr \cr 4x+2y=16 \end{cases}
Multiply the equations by a constant
Multiplying 4x+2y=16 by 2 makes the coefficients opposite of each other:
\begin{cases} 3x-4y=12 \cr \cr 8x+4y=32 \end{cases}
Add the equations
Adding the two equations gives:
11x=44
Divide both sides by 11:
x=4
Substitute to solve for the other variable
Substituting x=4 into 3x-4y=12 :
3\left(4\right)-4y=12
12-4y=12
Subtract 12 from both sides:
-4y=0
Divide both sides by -4:
y=0
The solution of this system is \left(4{,}0\right).