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4x+2y=100,x+y=30
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
4x+2y=100
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x=-2y+100
Subtract 2y from both sides of the equation.
x=\frac{1}{4}\left(-2y+100\right)
Divide both sides by 4.
x=-\frac{1}{2}y+25
Multiply \frac{1}{4} times -2y+100.
-\frac{1}{2}y+25+y=30
Substitute -\frac{y}{2}+25 for x in the other equation, x+y=30.
\frac{1}{2}y+25=30
Add -\frac{y}{2} to y.
\frac{1}{2}y=5
Subtract 25 from both sides of the equation.
y=10
Multiply both sides by 2.
x=-\frac{1}{2}\times 10+25
Substitute 10 for y in x=-\frac{1}{2}y+25. Because the resulting equation contains only one variable, you can solve for x directly.
x=-5+25
Multiply -\frac{1}{2} times 10.
x=20
Add 25 to -5.
x=20,y=10
The system is now solved.
4x+2y=100,x+y=30
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\30\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&2\\1&1\end{matrix}\right))\left(\begin{matrix}4&2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\1&1\end{matrix}\right))\left(\begin{matrix}100\\30\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&2\\1&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\1&1\end{matrix}\right))\left(\begin{matrix}100\\30\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\1&1\end{matrix}\right))\left(\begin{matrix}100\\30\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4-2}&-\frac{2}{4-2}\\-\frac{1}{4-2}&\frac{4}{4-2}\end{matrix}\right)\left(\begin{matrix}100\\30\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-1\\-\frac{1}{2}&2\end{matrix}\right)\left(\begin{matrix}100\\30\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 100-30\\-\frac{1}{2}\times 100+2\times 30\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\10\end{matrix}\right)
Do the arithmetic.
x=20,y=10
Extract the matrix elements x and y.
4x+2y=100,x+y=30
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
4x+2y=100,4x+4y=4\times 30
To make 4x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 4.
4x+2y=100,4x+4y=120
Simplify.
4x-4x+2y-4y=100-120
Subtract 4x+4y=120 from 4x+2y=100 by subtracting like terms on each side of the equal sign.
2y-4y=100-120
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-2y=100-120
Add 2y to -4y.
-2y=-20
Add 100 to -120.
y=10
Divide both sides by -2.
x+10=30
Substitute 10 for y in x+y=30. Because the resulting equation contains only one variable, you can solve for x directly.
x=20
Subtract 10 from both sides of the equation.
x=20,y=10
The system is now solved.