\left\{ \begin{array} { l } { 5 x + 7 y = 50 } \\ { 9 x + 14 y = 97 } \end{array} \right\}
Solve for x, y
x=3
y=5
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5x+7y=50,9x+14y=97
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.
5x+7y=50
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=-7y+50
Subtract 7y from both sides of the equation.
x=\frac{1}{5}\left(-7y+50\right)
Divide both sides by 5.
x=-\frac{7}{5}y+10
Multiply \frac{1}{5} times -7y+50.
9\left(-\frac{7}{5}y+10\right)+14y=97
Substitute -\frac{7y}{5}+10 for x in the other equation, 9x+14y=97.
-\frac{63}{5}y+90+14y=97
Multiply 9 times -\frac{7y}{5}+10.
\frac{7}{5}y+90=97
Add -\frac{63y}{5} to 14y.
\frac{7}{5}y=7
Subtract 90 from both sides of the equation.
y=5
Divide both sides of the equation by \frac{7}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{7}{5}\times 5+10
Substitute 5 for y in x=-\frac{7}{5}y+10. Because the resulting equation contains only one variable, you can solve for x directly.
x=-7+10
Multiply -\frac{7}{5} times 5.
x=3
Add 10 to -7.
x=3,y=5
The system is now solved.
5x+7y=50,9x+14y=97
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&7\\9&14\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\97\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&7\\9&14\end{matrix}\right))\left(\begin{matrix}5&7\\9&14\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&7\\9&14\end{matrix}\right))\left(\begin{matrix}50\\97\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&7\\9&14\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}5&7\\9&14\end{matrix}\right))\left(\begin{matrix}50\\97\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}5&7\\9&14\end{matrix}\right))\left(\begin{matrix}50\\97\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{14}{5\times 14-7\times 9}&-\frac{7}{5\times 14-7\times 9}\\-\frac{9}{5\times 14-7\times 9}&\frac{5}{5\times 14-7\times 9}\end{matrix}\right)\left(\begin{matrix}50\\97\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&-1\\-\frac{9}{7}&\frac{5}{7}\end{matrix}\right)\left(\begin{matrix}50\\97\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 50-97\\-\frac{9}{7}\times 50+\frac{5}{7}\times 97\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\5\end{matrix}\right)
Do the arithmetic.
x=3,y=5
Extract the matrix elements x and y.
5x+7y=50,9x+14y=97
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.
9\times 5x+9\times 7y=9\times 50,5\times 9x+5\times 14y=5\times 97
To make 5x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 5.
45x+63y=450,45x+70y=485
Simplify.
45x-45x+63y-70y=450-485
Subtract 45x+70y=485 from 45x+63y=450 by subtracting like terms on each side of the equal sign.
63y-70y=450-485
Add 45x to -45x. Terms 45x and -45x cancel out, leaving an equation with only one variable that can be solved.
-7y=450-485
Add 63y to -70y.
-7y=-35
Add 450 to -485.
y=5
Divide both sides by -7.
9x+14\times 5=97
Substitute 5 for y in 9x+14y=97. Because the resulting equation contains only one variable, you can solve for x directly.
9x+70=97
Multiply 14 times 5.
9x=27
Subtract 70 from both sides of the equation.
x=3
Divide both sides by 9.
x=3,y=5
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}