Solve for x, y
x = \frac{80}{7} = 11\frac{3}{7} \approx 11.428571429
y = \frac{550}{7} = 78\frac{4}{7} \approx 78.571428571
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16x+2y=340,2x+2y=180
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.
16x+2y=340
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
16x=-2y+340
Subtract 2y from both sides of the equation.
x=\frac{1}{16}\left(-2y+340\right)
Divide both sides by 16.
x=-\frac{1}{8}y+\frac{85}{4}
Multiply \frac{1}{16} times -2y+340.
2\left(-\frac{1}{8}y+\frac{85}{4}\right)+2y=180
Substitute -\frac{y}{8}+\frac{85}{4} for x in the other equation, 2x+2y=180.
-\frac{1}{4}y+\frac{85}{2}+2y=180
Multiply 2 times -\frac{y}{8}+\frac{85}{4}.
\frac{7}{4}y+\frac{85}{2}=180
Add -\frac{y}{4} to 2y.
\frac{7}{4}y=\frac{275}{2}
Subtract \frac{85}{2} from both sides of the equation.
y=\frac{550}{7}
Divide both sides of the equation by \frac{7}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{8}\times \frac{550}{7}+\frac{85}{4}
Substitute \frac{550}{7} for y in x=-\frac{1}{8}y+\frac{85}{4}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{275}{28}+\frac{85}{4}
Multiply -\frac{1}{8} times \frac{550}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{80}{7}
Add \frac{85}{4} to -\frac{275}{28} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{80}{7},y=\frac{550}{7}
The system is now solved.
16x+2y=340,2x+2y=180
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}16&2\\2&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}340\\180\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}16&2\\2&2\end{matrix}\right))\left(\begin{matrix}16&2\\2&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}16&2\\2&2\end{matrix}\right))\left(\begin{matrix}340\\180\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}16&2\\2&2\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}16&2\\2&2\end{matrix}\right))\left(\begin{matrix}340\\180\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}16&2\\2&2\end{matrix}\right))\left(\begin{matrix}340\\180\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{2}{16\times 2-2\times 2}&-\frac{2}{16\times 2-2\times 2}\\-\frac{2}{16\times 2-2\times 2}&\frac{16}{16\times 2-2\times 2}\end{matrix}\right)\left(\begin{matrix}340\\180\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}{14}&-\frac{1}{14}\\-\frac{1}{14}&\frac{4}{7}\end{matrix}\right)\left(\begin{matrix}340\\180\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{14}\times 340-\frac{1}{14}\times 180\\-\frac{1}{14}\times 340+\frac{4}{7}\times 180\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{7}\\\frac{550}{7}\end{matrix}\right)
Do the arithmetic.
x=\frac{80}{7},y=\frac{550}{7}
Extract the matrix elements x and y.
16x+2y=340,2x+2y=180
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.
16x-2x+2y-2y=340-180
Subtract 2x+2y=180 from 16x+2y=340 by subtracting like terms on each side of the equal sign.
16x-2x=340-180
Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.
14x=340-180
Add 16x to -2x.
14x=160
Add 340 to -180.
x=\frac{80}{7}
Divide both sides by 14.
2\times \frac{80}{7}+2y=180
Substitute \frac{80}{7} for x in 2x+2y=180. Because the resulting equation contains only one variable, you can solve for y directly.
\frac{160}{7}+2y=180
Multiply 2 times \frac{80}{7}.
2y=\frac{1100}{7}
Subtract \frac{160}{7} from both sides of the equation.
y=\frac{550}{7}
Divide both sides by 2.
x=\frac{80}{7},y=\frac{550}{7}
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
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Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
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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}