Solve for x, y

x=5<br/>y=20

$x=5$

$y=20$

$y=20$

Steps Using Substitution

Steps Using Matrices

Steps Using Elimination

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x+15-y=0

Consider the first equation. Subtract y from both sides.

x-y=-15

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

4x-y=0

Consider the second equation. Subtract y from both sides.

x-y=-15,4x-y=0

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-y=-15

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

x=y-15

Add y to both sides of the equation.

4\left(y-15\right)-y=0

Substitute y-15 for x in the other equation, 4x-y=0.

4y-60-y=0

Multiply 4 times y-15.

3y-60=0

Add 4y to -y.

3y=60

Add 60 to both sides of the equation.

y=20

Divide both sides by 3.

x=20-15

Substitute 20 for y in x=y-15. Because the resulting equation contains only one variable, you can solve for x directly.

x=5

Add -15 to 20.

x=5,y=20

The system is now solved.

x+15-y=0

Consider the first equation. Subtract y from both sides.

x-y=-15

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

4x-y=0

Consider the second equation. Subtract y from both sides.

x-y=-15,4x-y=0

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

\left(\begin{matrix}1&-1\\4&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-15\\0\end{matrix}\right)

Write the equations in matrix form.

inverse(\left(\begin{matrix}1&-1\\4&-1\end{matrix}\right))\left(\begin{matrix}1&-1\\4&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\4&-1\end{matrix}\right))\left(\begin{matrix}-15\\0\end{matrix}\right)

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\4&-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}1&-1\\4&-1\end{matrix}\right))\left(\begin{matrix}-15\\0\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&-1\\4&-1\end{matrix}\right))\left(\begin{matrix}-15\\0\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}{-1-\left(-4\right)}&-\frac{-1}{-1-\left(-4\right)}\\-\frac{4}{-1-\left(-4\right)}&\frac{1}{-1-\left(-4\right)}\end{matrix}\right)\left(\begin{matrix}-15\\0\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}{3}\\-\frac{4}{3}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-15\\0\end{matrix}\right)

Do the arithmetic.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\left(-15\right)\\-\frac{4}{3}\left(-15\right)\end{matrix}\right)

Multiply the matrices.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\20\end{matrix}\right)

Do the arithmetic.

x=5,y=20

Extract the matrix elements x and y.

x+15-y=0

Consider the first equation. Subtract y from both sides.

x-y=-15

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

4x-y=0

Consider the second equation. Subtract y from both sides.

x-y=-15,4x-y=0

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.

x-4x-y+y=-15

Subtract 4x-y=0 from x-y=-15 by subtracting like terms on each side of the equal sign.

x-4x=-15

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

-3x=-15

Add x to -4x.

x=5

Divide both sides by -3.

4\times 5-y=0

Substitute 5 for x in 4x-y=0. Because the resulting equation contains only one variable, you can solve for y directly.

20-y=0

Multiply 4 times 5.

-y=-20

Subtract 20 from both sides of the equation.

y=20

Divide both sides by -1.

x=5,y=20

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 $