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x-3y=38,4x+6y=26
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-3y=38
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=3y+38
Add 3y to both sides of the equation.
4\left(3y+38\right)+6y=26
Substitute 3y+38 for x in the other equation, 4x+6y=26.
12y+152+6y=26
Multiply 4 times 3y+38.
18y+152=26
Add 12y to 6y.
18y=-126
Subtract 152 from both sides of the equation.
y=-7
Divide both sides by 18.
x=3\left(-7\right)+38
Substitute -7 for y in x=3y+38. Because the resulting equation contains only one variable, you can solve for x directly.
x=-21+38
Multiply 3 times -7.
x=17
Add 38 to -21.
x=17,y=-7
The system is now solved.
x-3y=38,4x+6y=26
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-3\\4&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}38\\26\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-3\\4&6\end{matrix}\right))\left(\begin{matrix}1&-3\\4&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\4&6\end{matrix}\right))\left(\begin{matrix}38\\26\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-3\\4&6\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&-3\\4&6\end{matrix}\right))\left(\begin{matrix}38\\26\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&-3\\4&6\end{matrix}\right))\left(\begin{matrix}38\\26\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{6}{6-\left(-3\times 4\right)}&-\frac{-3}{6-\left(-3\times 4\right)}\\-\frac{4}{6-\left(-3\times 4\right)}&\frac{1}{6-\left(-3\times 4\right)}\end{matrix}\right)\left(\begin{matrix}38\\26\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}{3}&\frac{1}{6}\\-\frac{2}{9}&\frac{1}{18}\end{matrix}\right)\left(\begin{matrix}38\\26\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 38+\frac{1}{6}\times 26\\-\frac{2}{9}\times 38+\frac{1}{18}\times 26\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}17\\-7\end{matrix}\right)
Do the arithmetic.
x=17,y=-7
Extract the matrix elements x and y.
x-3y=38,4x+6y=26
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+4\left(-3\right)y=4\times 38,4x+6y=26
To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.
4x-12y=152,4x+6y=26
Simplify.
4x-4x-12y-6y=152-26
Subtract 4x+6y=26 from 4x-12y=152 by subtracting like terms on each side of the equal sign.
-12y-6y=152-26
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-18y=152-26
Add -12y to -6y.
-18y=126
Add 152 to -26.
y=-7
Divide both sides by -18.
4x+6\left(-7\right)=26
Substitute -7 for y in 4x+6y=26. Because the resulting equation contains only one variable, you can solve for x directly.
4x-42=26
Multiply 6 times -7.
4x=68
Add 42 to both sides of the equation.
x=17
Divide both sides by 4.
x=17,y=-7
The system is now solved.