\left\{ \begin{array} { l } { \frac { 2 x } { 3 } - y + 1 = 2 x - \frac { 2 y } { 5 } } \\ { \frac { x } { 4 } = 8 + \frac { y } { 5 } } \end{array} \right.
Solve for x, y
x=12
y=-25
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5\times 2x-15y+15=30x-3\times 2y
Consider the first equation. Multiply both sides of the equation by 15, the least common multiple of 3,5.
10x-15y+15=30x-3\times 2y
Multiply 5 and 2 to get 10.
10x-15y+15=30x-6y
Multiply -3 and 2 to get -6.
10x-15y+15-30x=-6y
Subtract 30x from both sides.
-20x-15y+15=-6y
Combine 10x and -30x to get -20x.
-20x-15y+15+6y=0
Add 6y to both sides.
-20x-9y+15=0
Combine -15y and 6y to get -9y.
-20x-9y=-15
Subtract 15 from both sides. Anything subtracted from zero gives its negation.
5x=160+4y
Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.
5x-4y=160
Subtract 4y from both sides.
-20x-9y=-15,5x-4y=160
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.
-20x-9y=-15
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
-20x=9y-15
Add 9y to both sides of the equation.
x=-\frac{1}{20}\left(9y-15\right)
Divide both sides by -20.
x=-\frac{9}{20}y+\frac{3}{4}
Multiply -\frac{1}{20} times 9y-15.
5\left(-\frac{9}{20}y+\frac{3}{4}\right)-4y=160
Substitute -\frac{9y}{20}+\frac{3}{4} for x in the other equation, 5x-4y=160.
-\frac{9}{4}y+\frac{15}{4}-4y=160
Multiply 5 times -\frac{9y}{20}+\frac{3}{4}.
-\frac{25}{4}y+\frac{15}{4}=160
Add -\frac{9y}{4} to -4y.
-\frac{25}{4}y=\frac{625}{4}
Subtract \frac{15}{4} from both sides of the equation.
y=-25
Divide both sides of the equation by -\frac{25}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{9}{20}\left(-25\right)+\frac{3}{4}
Substitute -25 for y in x=-\frac{9}{20}y+\frac{3}{4}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{45+3}{4}
Multiply -\frac{9}{20} times -25.
x=12
Add \frac{3}{4} to \frac{45}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=12,y=-25
The system is now solved.
5\times 2x-15y+15=30x-3\times 2y
Consider the first equation. Multiply both sides of the equation by 15, the least common multiple of 3,5.
10x-15y+15=30x-3\times 2y
Multiply 5 and 2 to get 10.
10x-15y+15=30x-6y
Multiply -3 and 2 to get -6.
10x-15y+15-30x=-6y
Subtract 30x from both sides.
-20x-15y+15=-6y
Combine 10x and -30x to get -20x.
-20x-15y+15+6y=0
Add 6y to both sides.
-20x-9y+15=0
Combine -15y and 6y to get -9y.
-20x-9y=-15
Subtract 15 from both sides. Anything subtracted from zero gives its negation.
5x=160+4y
Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.
5x-4y=160
Subtract 4y from both sides.
-20x-9y=-15,5x-4y=160
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}-20&-9\\5&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-15\\160\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}-20&-9\\5&-4\end{matrix}\right))\left(\begin{matrix}-20&-9\\5&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-20&-9\\5&-4\end{matrix}\right))\left(\begin{matrix}-15\\160\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}-20&-9\\5&-4\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}-20&-9\\5&-4\end{matrix}\right))\left(\begin{matrix}-15\\160\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}-20&-9\\5&-4\end{matrix}\right))\left(\begin{matrix}-15\\160\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{4}{-20\left(-4\right)-\left(-9\times 5\right)}&-\frac{-9}{-20\left(-4\right)-\left(-9\times 5\right)}\\-\frac{5}{-20\left(-4\right)-\left(-9\times 5\right)}&-\frac{20}{-20\left(-4\right)-\left(-9\times 5\right)}\end{matrix}\right)\left(\begin{matrix}-15\\160\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{4}{125}&\frac{9}{125}\\-\frac{1}{25}&-\frac{4}{25}\end{matrix}\right)\left(\begin{matrix}-15\\160\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{125}\left(-15\right)+\frac{9}{125}\times 160\\-\frac{1}{25}\left(-15\right)-\frac{4}{25}\times 160\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\-25\end{matrix}\right)
Do the arithmetic.
x=12,y=-25
Extract the matrix elements x and y.
5\times 2x-15y+15=30x-3\times 2y
Consider the first equation. Multiply both sides of the equation by 15, the least common multiple of 3,5.
10x-15y+15=30x-3\times 2y
Multiply 5 and 2 to get 10.
10x-15y+15=30x-6y
Multiply -3 and 2 to get -6.
10x-15y+15-30x=-6y
Subtract 30x from both sides.
-20x-15y+15=-6y
Combine 10x and -30x to get -20x.
-20x-15y+15+6y=0
Add 6y to both sides.
-20x-9y+15=0
Combine -15y and 6y to get -9y.
-20x-9y=-15
Subtract 15 from both sides. Anything subtracted from zero gives its negation.
5x=160+4y
Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.
5x-4y=160
Subtract 4y from both sides.
-20x-9y=-15,5x-4y=160
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.
5\left(-20\right)x+5\left(-9\right)y=5\left(-15\right),-20\times 5x-20\left(-4\right)y=-20\times 160
To make -20x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by -20.
-100x-45y=-75,-100x+80y=-3200
Simplify.
-100x+100x-45y-80y=-75+3200
Subtract -100x+80y=-3200 from -100x-45y=-75 by subtracting like terms on each side of the equal sign.
-45y-80y=-75+3200
Add -100x to 100x. Terms -100x and 100x cancel out, leaving an equation with only one variable that can be solved.
-125y=-75+3200
Add -45y to -80y.
-125y=3125
Add -75 to 3200.
y=-25
Divide both sides by -125.
5x-4\left(-25\right)=160
Substitute -25 for y in 5x-4y=160. Because the resulting equation contains only one variable, you can solve for x directly.
5x+100=160
Multiply -4 times -25.
5x=60
Subtract 100 from both sides of the equation.
x=12
Divide both sides by 5.
x=12,y=-25
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
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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}