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x+4y=900,3x-4y=300
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.
x+4y=900
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-4y+900
Subtract 4y from both sides of the equation.
3\left(-4y+900\right)-4y=300
Substitute -4y+900 for x in the other equation, 3x-4y=300.
-12y+2700-4y=300
Multiply 3 times -4y+900.
-16y+2700=300
Add -12y to -4y.
-16y=-2400
Subtract 2700 from both sides of the equation.
y=150
Divide both sides by -16.
x=-4\times 150+900
Substitute 150 for y in x=-4y+900. Because the resulting equation contains only one variable, you can solve for x directly.
x=-600+900
Multiply -4 times 150.
x=300
Add 900 to -600.
x=300,y=150
The system is now solved.
x+4y=900,3x-4y=300
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&4\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}900\\300\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&4\\3&-4\end{matrix}\right))\left(\begin{matrix}1&4\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&4\\3&-4\end{matrix}\right))\left(\begin{matrix}900\\300\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&4\\3&-4\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}1&4\\3&-4\end{matrix}\right))\left(\begin{matrix}900\\300\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}1&4\\3&-4\end{matrix}\right))\left(\begin{matrix}900\\300\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{4}{-4-4\times 3}&-\frac{4}{-4-4\times 3}\\-\frac{3}{-4-4\times 3}&\frac{1}{-4-4\times 3}\end{matrix}\right)\left(\begin{matrix}900\\300\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}{4}&\frac{1}{4}\\\frac{3}{16}&-\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}900\\300\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\times 900+\frac{1}{4}\times 300\\\frac{3}{16}\times 900-\frac{1}{16}\times 300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\150\end{matrix}\right)
Do the arithmetic.
x=300,y=150
Extract the matrix elements x and y.
x+4y=900,3x-4y=300
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.
3x+3\times 4y=3\times 900,3x-4y=300
To make x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.
3x+12y=2700,3x-4y=300
Simplify.
3x-3x+12y+4y=2700-300
Subtract 3x-4y=300 from 3x+12y=2700 by subtracting like terms on each side of the equal sign.
12y+4y=2700-300
Add 3x to -3x. Terms 3x and -3x cancel out, leaving an equation with only one variable that can be solved.
16y=2700-300
Add 12y to 4y.
16y=2400
Add 2700 to -300.
y=150
Divide both sides by 16.
3x-4\times 150=300
Substitute 150 for y in 3x-4y=300. Because the resulting equation contains only one variable, you can solve for x directly.
3x-600=300
Multiply -4 times 150.
3x=900
Add 600 to both sides of the equation.
x=300
Divide both sides by 3.
x=300,y=150
The system is now solved.