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4x+3y=25,x+y=45
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+3y=25
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x=-3y+25
Subtract 3y from both sides of the equation.
x=\frac{1}{4}\left(-3y+25\right)
Divide both sides by 4.
x=-\frac{3}{4}y+\frac{25}{4}
Multiply \frac{1}{4} times -3y+25.
-\frac{3}{4}y+\frac{25}{4}+y=45
Substitute \frac{-3y+25}{4} for x in the other equation, x+y=45.
\frac{1}{4}y+\frac{25}{4}=45
Add -\frac{3y}{4} to y.
\frac{1}{4}y=\frac{155}{4}
Subtract \frac{25}{4} from both sides of the equation.
y=155
Multiply both sides by 4.
x=-\frac{3}{4}\times 155+\frac{25}{4}
Substitute 155 for y in x=-\frac{3}{4}y+\frac{25}{4}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-465+25}{4}
Multiply -\frac{3}{4} times 155.
x=-110
Add \frac{25}{4} to -\frac{465}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=-110,y=155
The system is now solved.
4x+3y=25,x+y=45
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&3\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\45\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&3\\1&1\end{matrix}\right))\left(\begin{matrix}4&3\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&3\\1&1\end{matrix}\right))\left(\begin{matrix}25\\45\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&3\\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&3\\1&1\end{matrix}\right))\left(\begin{matrix}25\\45\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&3\\1&1\end{matrix}\right))\left(\begin{matrix}25\\45\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}{4-3}&-\frac{3}{4-3}\\-\frac{1}{4-3}&\frac{4}{4-3}\end{matrix}\right)\left(\begin{matrix}25\\45\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}1&-3\\-1&4\end{matrix}\right)\left(\begin{matrix}25\\45\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25-3\times 45\\-25+4\times 45\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-110\\155\end{matrix}\right)
Do the arithmetic.
x=-110,y=155
Extract the matrix elements x and y.
4x+3y=25,x+y=45
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+3y=25,4x+4y=4\times 45
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+3y=25,4x+4y=180
Simplify.
4x-4x+3y-4y=25-180
Subtract 4x+4y=180 from 4x+3y=25 by subtracting like terms on each side of the equal sign.
3y-4y=25-180
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-y=25-180
Add 3y to -4y.
-y=-155
Add 25 to -180.
y=155
Divide both sides by -1.
x+155=45
Substitute 155 for y in x+y=45. Because the resulting equation contains only one variable, you can solve for x directly.
x=-110
Subtract 155 from both sides of the equation.
x=-110,y=155
The system is now solved.