\left\{ \begin{array} { l } { x + y = 5 } \\ { \frac { x } { 0.2 } + \frac { y } { 0.1 } = 470 } \end{array} \right.
Solve for x, y
x=-84
y=89
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x+y=5,5x+10y=470
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=5
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+5
Subtract y from both sides of the equation.
5\left(-y+5\right)+10y=470
Substitute -y+5 for x in the other equation, 5x+10y=470.
-5y+25+10y=470
Multiply 5 times -y+5.
5y+25=470
Add -5y to 10y.
5y=445
Subtract 25 from both sides of the equation.
y=89
Divide both sides by 5.
x=-89+5
Substitute 89 for y in x=-y+5. Because the resulting equation contains only one variable, you can solve for x directly.
x=-84
Add 5 to -89.
x=-84,y=89
The system is now solved.
x+y=5,5x+10y=470
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\5&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\470\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\5&10\end{matrix}\right))\left(\begin{matrix}1&1\\5&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\5&10\end{matrix}\right))\left(\begin{matrix}5\\470\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\5&10\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\\5&10\end{matrix}\right))\left(\begin{matrix}5\\470\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\\5&10\end{matrix}\right))\left(\begin{matrix}5\\470\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{10}{10-5}&-\frac{1}{10-5}\\-\frac{5}{10-5}&\frac{1}{10-5}\end{matrix}\right)\left(\begin{matrix}5\\470\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}2&-\frac{1}{5}\\-1&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}5\\470\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 5-\frac{1}{5}\times 470\\-5+\frac{1}{5}\times 470\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-84\\89\end{matrix}\right)
Do the arithmetic.
x=-84,y=89
Extract the matrix elements x and y.
x+y=5,5x+10y=470
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.
5x+5y=5\times 5,5x+10y=470
To make x 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 1.
5x+5y=25,5x+10y=470
Simplify.
5x-5x+5y-10y=25-470
Subtract 5x+10y=470 from 5x+5y=25 by subtracting like terms on each side of the equal sign.
5y-10y=25-470
Add 5x to -5x. Terms 5x and -5x cancel out, leaving an equation with only one variable that can be solved.
-5y=25-470
Add 5y to -10y.
-5y=-445
Add 25 to -470.
y=89
Divide both sides by -5.
5x+10\times 89=470
Substitute 89 for y in 5x+10y=470. Because the resulting equation contains only one variable, you can solve for x directly.
5x+890=470
Multiply 10 times 89.
5x=-420
Subtract 890 from both sides of the equation.
x=-84
Divide both sides by 5.
x=-84,y=89
The system is now solved.
Examples
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{ 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}