\left\{ \begin{array} { l } { 4 x + y = 20 } \\ { x + 5 y = 25 } \end{array} \right.
Solve for x, y
x = \frac{75}{19} = 3\frac{18}{19} \approx 3.947368421
y = \frac{80}{19} = 4\frac{4}{19} \approx 4.210526316
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4x+y=20,x+5y=25
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.
4x+y=20
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x=-y+20
Subtract y from both sides of the equation.
x=\frac{1}{4}\left(-y+20\right)
Divide both sides by 4.
x=-\frac{1}{4}y+5
Multiply \frac{1}{4} times -y+20.
-\frac{1}{4}y+5+5y=25
Substitute -\frac{y}{4}+5 for x in the other equation, x+5y=25.
\frac{19}{4}y+5=25
Add -\frac{y}{4} to 5y.
\frac{19}{4}y=20
Subtract 5 from both sides of the equation.
y=\frac{80}{19}
Divide both sides of the equation by \frac{19}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{4}\times \frac{80}{19}+5
Substitute \frac{80}{19} for y in x=-\frac{1}{4}y+5. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{20}{19}+5
Multiply -\frac{1}{4} times \frac{80}{19} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{75}{19}
Add 5 to -\frac{20}{19}.
x=\frac{75}{19},y=\frac{80}{19}
The system is now solved.
4x+y=20,x+5y=25
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&1\\1&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\25\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&1\\1&5\end{matrix}\right))\left(\begin{matrix}4&1\\1&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\1&5\end{matrix}\right))\left(\begin{matrix}20\\25\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&1\\1&5\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}4&1\\1&5\end{matrix}\right))\left(\begin{matrix}20\\25\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}4&1\\1&5\end{matrix}\right))\left(\begin{matrix}20\\25\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{5}{4\times 5-1}&-\frac{1}{4\times 5-1}\\-\frac{1}{4\times 5-1}&\frac{4}{4\times 5-1}\end{matrix}\right)\left(\begin{matrix}20\\25\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{5}{19}&-\frac{1}{19}\\-\frac{1}{19}&\frac{4}{19}\end{matrix}\right)\left(\begin{matrix}20\\25\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}\times 20-\frac{1}{19}\times 25\\-\frac{1}{19}\times 20+\frac{4}{19}\times 25\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{75}{19}\\\frac{80}{19}\end{matrix}\right)
Do the arithmetic.
x=\frac{75}{19},y=\frac{80}{19}
Extract the matrix elements x and y.
4x+y=20,x+5y=25
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.
4x+y=20,4x+4\times 5y=4\times 25
To make 4x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 4.
4x+y=20,4x+20y=100
Simplify.
4x-4x+y-20y=20-100
Subtract 4x+20y=100 from 4x+y=20 by subtracting like terms on each side of the equal sign.
y-20y=20-100
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-19y=20-100
Add y to -20y.
-19y=-80
Add 20 to -100.
y=\frac{80}{19}
Divide both sides by -19.
x+5\times \frac{80}{19}=25
Substitute \frac{80}{19} for y in x+5y=25. Because the resulting equation contains only one variable, you can solve for x directly.
x+\frac{400}{19}=25
Multiply 5 times \frac{80}{19}.
x=\frac{75}{19}
Subtract \frac{400}{19} from both sides of the equation.
x=\frac{75}{19},y=\frac{80}{19}
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}