Solve for x, y
x = \frac{5}{4} = 1\frac{1}{4} = 1.25
y=\frac{1}{2}=0.5
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40x+56y=78,24x+40y=50
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.
40x+56y=78
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
40x=-56y+78
Subtract 56y from both sides of the equation.
x=\frac{1}{40}\left(-56y+78\right)
Divide both sides by 40.
x=-\frac{7}{5}y+\frac{39}{20}
Multiply \frac{1}{40} times -56y+78.
24\left(-\frac{7}{5}y+\frac{39}{20}\right)+40y=50
Substitute -\frac{7y}{5}+\frac{39}{20} for x in the other equation, 24x+40y=50.
-\frac{168}{5}y+\frac{234}{5}+40y=50
Multiply 24 times -\frac{7y}{5}+\frac{39}{20}.
\frac{32}{5}y+\frac{234}{5}=50
Add -\frac{168y}{5} to 40y.
\frac{32}{5}y=\frac{16}{5}
Subtract \frac{234}{5} from both sides of the equation.
y=\frac{1}{2}
Divide both sides of the equation by \frac{32}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{7}{5}\times \frac{1}{2}+\frac{39}{20}
Substitute \frac{1}{2} for y in x=-\frac{7}{5}y+\frac{39}{20}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{7}{10}+\frac{39}{20}
Multiply -\frac{7}{5} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{5}{4}
Add \frac{39}{20} to -\frac{7}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{5}{4},y=\frac{1}{2}
The system is now solved.
40x+56y=78,24x+40y=50
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}40&56\\24&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}78\\50\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}40&56\\24&40\end{matrix}\right))\left(\begin{matrix}40&56\\24&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&56\\24&40\end{matrix}\right))\left(\begin{matrix}78\\50\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}40&56\\24&40\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}40&56\\24&40\end{matrix}\right))\left(\begin{matrix}78\\50\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}40&56\\24&40\end{matrix}\right))\left(\begin{matrix}78\\50\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{40}{40\times 40-56\times 24}&-\frac{56}{40\times 40-56\times 24}\\-\frac{24}{40\times 40-56\times 24}&\frac{40}{40\times 40-56\times 24}\end{matrix}\right)\left(\begin{matrix}78\\50\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}{32}&-\frac{7}{32}\\-\frac{3}{32}&\frac{5}{32}\end{matrix}\right)\left(\begin{matrix}78\\50\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{32}\times 78-\frac{7}{32}\times 50\\-\frac{3}{32}\times 78+\frac{5}{32}\times 50\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{4}\\\frac{1}{2}\end{matrix}\right)
Do the arithmetic.
x=\frac{5}{4},y=\frac{1}{2}
Extract the matrix elements x and y.
40x+56y=78,24x+40y=50
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.
24\times 40x+24\times 56y=24\times 78,40\times 24x+40\times 40y=40\times 50
To make 40x and 24x equal, multiply all terms on each side of the first equation by 24 and all terms on each side of the second by 40.
960x+1344y=1872,960x+1600y=2000
Simplify.
960x-960x+1344y-1600y=1872-2000
Subtract 960x+1600y=2000 from 960x+1344y=1872 by subtracting like terms on each side of the equal sign.
1344y-1600y=1872-2000
Add 960x to -960x. Terms 960x and -960x cancel out, leaving an equation with only one variable that can be solved.
-256y=1872-2000
Add 1344y to -1600y.
-256y=-128
Add 1872 to -2000.
y=\frac{1}{2}
Divide both sides by -256.
24x+40\times \frac{1}{2}=50
Substitute \frac{1}{2} for y in 24x+40y=50. Because the resulting equation contains only one variable, you can solve for x directly.
24x+20=50
Multiply 40 times \frac{1}{2}.
24x=30
Subtract 20 from both sides of the equation.
x=\frac{5}{4}
Divide both sides by 24.
x=\frac{5}{4},y=\frac{1}{2}
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}