\left\{ \begin{array} { l } { 3 x + 5 y = 165 } \\ { 4 x + 7 y = 225 } \end{array} \right.
Solve for x, y
x=30
y=15
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3x+5y=165,4x+7y=225
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.
3x+5y=165
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-5y+165
Subtract 5y from both sides of the equation.
x=\frac{1}{3}\left(-5y+165\right)
Divide both sides by 3.
x=-\frac{5}{3}y+55
Multiply \frac{1}{3} times -5y+165.
4\left(-\frac{5}{3}y+55\right)+7y=225
Substitute -\frac{5y}{3}+55 for x in the other equation, 4x+7y=225.
-\frac{20}{3}y+220+7y=225
Multiply 4 times -\frac{5y}{3}+55.
\frac{1}{3}y+220=225
Add -\frac{20y}{3} to 7y.
\frac{1}{3}y=5
Subtract 220 from both sides of the equation.
y=15
Multiply both sides by 3.
x=-\frac{5}{3}\times 15+55
Substitute 15 for y in x=-\frac{5}{3}y+55. Because the resulting equation contains only one variable, you can solve for x directly.
x=-25+55
Multiply -\frac{5}{3} times 15.
x=30
Add 55 to -25.
x=30,y=15
The system is now solved.
3x+5y=165,4x+7y=225
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&5\\4&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}165\\225\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&5\\4&7\end{matrix}\right))\left(\begin{matrix}3&5\\4&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&5\\4&7\end{matrix}\right))\left(\begin{matrix}165\\225\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&5\\4&7\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}3&5\\4&7\end{matrix}\right))\left(\begin{matrix}165\\225\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}3&5\\4&7\end{matrix}\right))\left(\begin{matrix}165\\225\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{7}{3\times 7-5\times 4}&-\frac{5}{3\times 7-5\times 4}\\-\frac{4}{3\times 7-5\times 4}&\frac{3}{3\times 7-5\times 4}\end{matrix}\right)\left(\begin{matrix}165\\225\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}7&-5\\-4&3\end{matrix}\right)\left(\begin{matrix}165\\225\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\times 165-5\times 225\\-4\times 165+3\times 225\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}30\\15\end{matrix}\right)
Do the arithmetic.
x=30,y=15
Extract the matrix elements x and y.
3x+5y=165,4x+7y=225
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.
4\times 3x+4\times 5y=4\times 165,3\times 4x+3\times 7y=3\times 225
To make 3x 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 3.
12x+20y=660,12x+21y=675
Simplify.
12x-12x+20y-21y=660-675
Subtract 12x+21y=675 from 12x+20y=660 by subtracting like terms on each side of the equal sign.
20y-21y=660-675
Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.
-y=660-675
Add 20y to -21y.
-y=-15
Add 660 to -675.
y=15
Divide both sides by -1.
4x+7\times 15=225
Substitute 15 for y in 4x+7y=225. Because the resulting equation contains only one variable, you can solve for x directly.
4x+105=225
Multiply 7 times 15.
4x=120
Subtract 105 from both sides of the equation.
x=30
Divide both sides by 4.
x=30,y=15
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}