\left\{ \begin{array} { l } { 3 x - 2 y = 10 } \\ { 3 y = + 4 x + 17 } \end{array} \right.
Solve for x, y
x=64
y=91
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3y-4x=17
Consider the second equation. Subtract 4x from both sides.
3x-2y=10,-4x+3y=17
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-2y=10
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=2y+10
Add 2y to both sides of the equation.
x=\frac{1}{3}\left(2y+10\right)
Divide both sides by 3.
x=\frac{2}{3}y+\frac{10}{3}
Multiply \frac{1}{3} times 10+2y.
-4\left(\frac{2}{3}y+\frac{10}{3}\right)+3y=17
Substitute \frac{10+2y}{3} for x in the other equation, -4x+3y=17.
-\frac{8}{3}y-\frac{40}{3}+3y=17
Multiply -4 times \frac{10+2y}{3}.
\frac{1}{3}y-\frac{40}{3}=17
Add -\frac{8y}{3} to 3y.
\frac{1}{3}y=\frac{91}{3}
Add \frac{40}{3} to both sides of the equation.
y=91
Multiply both sides by 3.
x=\frac{2}{3}\times 91+\frac{10}{3}
Substitute 91 for y in x=\frac{2}{3}y+\frac{10}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{182+10}{3}
Multiply \frac{2}{3} times 91.
x=64
Add \frac{10}{3} to \frac{182}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=64,y=91
The system is now solved.
3y-4x=17
Consider the second equation. Subtract 4x from both sides.
3x-2y=10,-4x+3y=17
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&-2\\-4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\17\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&-2\\-4&3\end{matrix}\right))\left(\begin{matrix}3&-2\\-4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\-4&3\end{matrix}\right))\left(\begin{matrix}10\\17\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-2\\-4&3\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&-2\\-4&3\end{matrix}\right))\left(\begin{matrix}10\\17\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&-2\\-4&3\end{matrix}\right))\left(\begin{matrix}10\\17\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{3}{3\times 3-\left(-2\left(-4\right)\right)}&-\frac{-2}{3\times 3-\left(-2\left(-4\right)\right)}\\-\frac{-4}{3\times 3-\left(-2\left(-4\right)\right)}&\frac{3}{3\times 3-\left(-2\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}10\\17\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}3&2\\4&3\end{matrix}\right)\left(\begin{matrix}10\\17\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\times 10+2\times 17\\4\times 10+3\times 17\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}64\\91\end{matrix}\right)
Do the arithmetic.
x=64,y=91
Extract the matrix elements x and y.
3y-4x=17
Consider the second equation. Subtract 4x from both sides.
3x-2y=10,-4x+3y=17
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\left(-2\right)y=-4\times 10,3\left(-4\right)x+3\times 3y=3\times 17
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+8y=-40,-12x+9y=51
Simplify.
-12x+12x+8y-9y=-40-51
Subtract -12x+9y=51 from -12x+8y=-40 by subtracting like terms on each side of the equal sign.
8y-9y=-40-51
Add -12x to 12x. Terms -12x and 12x cancel out, leaving an equation with only one variable that can be solved.
-y=-40-51
Add 8y to -9y.
-y=-91
Add -40 to -51.
y=91
Divide both sides by -1.
-4x+3\times 91=17
Substitute 91 for y in -4x+3y=17. Because the resulting equation contains only one variable, you can solve for x directly.
-4x+273=17
Multiply 3 times 91.
-4x=-256
Subtract 273 from both sides of the equation.
x=64
Divide both sides by -4.
x=64,y=91
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}