Solve for x, y

x=-4<br/>y=6

$x=−4$

$y=6$

$y=6$

Steps Using Substitution

Steps Using Matrices

Steps Using Elimination

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2x+3y=10,-3x+y=18

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.

2x+3y=10

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

2x=-3y+10

Subtract 3y from both sides of the equation.

x=\frac{1}{2}\left(-3y+10\right)

Divide both sides by 2.

x=-\frac{3}{2}y+5

Multiply \frac{1}{2}=0.5 times -3y+10.

-3\left(-\frac{3}{2}y+5\right)+y=18

Substitute -\frac{3y}{2}+5 for x in the other equation, -3x+y=18.

\frac{9}{2}y-15+y=18

Multiply -3 times -\frac{3y}{2}+5.

\frac{11}{2}y-15=18

Add \frac{9y}{2} to y.

\frac{11}{2}y=33

Add 15 to both sides of the equation.

y=6

Divide both sides of the equation by \frac{11}{2}=5.5, which is the same as multiplying both sides by the reciprocal of the fraction.

x=-\frac{3}{2}\times 6+5

Substitute 6 for y in x=-\frac{3}{2}y+5. Because the resulting equation contains only one variable, you can solve for x directly.

x=-9+5

Multiply -\frac{3}{2}=-1.5 times 6.

x=-4

Add 5 to -9.

x=-4,y=6

The system is now solved.

2x+3y=10,-3x+y=18

Put the equations in standard form and then use matrices to solve the system of equations.

\left(\begin{matrix}2&3\\-3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\18\end{matrix}\right)

Write the equations in matrix form.

inverse(\left(\begin{matrix}2&3\\-3&1\end{matrix}\right))\left(\begin{matrix}2&3\\-3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\-3&1\end{matrix}\right))\left(\begin{matrix}10\\18\end{matrix}\right)

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\-3&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}2&3\\-3&1\end{matrix}\right))\left(\begin{matrix}10\\18\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}2&3\\-3&1\end{matrix}\right))\left(\begin{matrix}10\\18\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}{2-3\left(-3\right)}&-\frac{3}{2-3\left(-3\right)}\\-\frac{-3}{2-3\left(-3\right)}&\frac{2}{2-3\left(-3\right)}\end{matrix}\right)\left(\begin{matrix}10\\18\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}{11}&-\frac{3}{11}\\\frac{3}{11}&\frac{2}{11}\end{matrix}\right)\left(\begin{matrix}10\\18\end{matrix}\right)

Do the arithmetic.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{11}\times 10-\frac{3}{11}\times 18\\\frac{3}{11}\times 10+\frac{2}{11}\times 18\end{matrix}\right)

Multiply the matrices.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\\6\end{matrix}\right)

Do the arithmetic.

x=-4,y=6

Extract the matrix elements x and y.

2x+3y=10,-3x+y=18

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.

-3\times 2x-3\times 3y=-3\times 10,2\left(-3\right)x+2y=2\times 18

To make 2x 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 2.

-6x-9y=-30,-6x+2y=36

Simplify.

-6x+6x-9y-2y=-30-36

Subtract -6x+2y=36 from -6x-9y=-30 by subtracting like terms on each side of the equal sign.

-9y-2y=-30-36

Add -6x to 6x. Terms -6x and 6x cancel out, leaving an equation with only one variable that can be solved.

-11y=-30-36

Add -9y to -2y.

-11y=-66

Add -30 to -36.

y=6

Divide both sides by -11.

-3x+6=18

Substitute 6 for y in -3x+y=18. Because the resulting equation contains only one variable, you can solve for x directly.

-3x=12

Subtract 6 from both sides of the equation.

x=-4

Divide both sides by -3.

x=-4,y=6

The system is now solved.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

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]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $