Solve for x, y
x = \frac{31}{2} = 15\frac{1}{2} = 15.5
y = \frac{13}{2} = 6\frac{1}{2} = 6.5
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5x+3y=97,2x+12y=109
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+3y=97
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=-3y+97
Subtract 3y from both sides of the equation.
x=\frac{1}{5}\left(-3y+97\right)
Divide both sides by 5.
x=-\frac{3}{5}y+\frac{97}{5}
Multiply \frac{1}{5} times -3y+97.
2\left(-\frac{3}{5}y+\frac{97}{5}\right)+12y=109
Substitute \frac{-3y+97}{5} for x in the other equation, 2x+12y=109.
-\frac{6}{5}y+\frac{194}{5}+12y=109
Multiply 2 times \frac{-3y+97}{5}.
\frac{54}{5}y+\frac{194}{5}=109
Add -\frac{6y}{5} to 12y.
\frac{54}{5}y=\frac{351}{5}
Subtract \frac{194}{5} from both sides of the equation.
y=\frac{13}{2}
Divide both sides of the equation by \frac{54}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3}{5}\times \frac{13}{2}+\frac{97}{5}
Substitute \frac{13}{2} for y in x=-\frac{3}{5}y+\frac{97}{5}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{39}{10}+\frac{97}{5}
Multiply -\frac{3}{5} times \frac{13}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{31}{2}
Add \frac{97}{5} to -\frac{39}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{31}{2},y=\frac{13}{2}
The system is now solved.
5x+3y=97,2x+12y=109
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&3\\2&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}97\\109\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&3\\2&12\end{matrix}\right))\left(\begin{matrix}5&3\\2&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&3\\2&12\end{matrix}\right))\left(\begin{matrix}97\\109\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&3\\2&12\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&3\\2&12\end{matrix}\right))\left(\begin{matrix}97\\109\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&3\\2&12\end{matrix}\right))\left(\begin{matrix}97\\109\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{12}{5\times 12-3\times 2}&-\frac{3}{5\times 12-3\times 2}\\-\frac{2}{5\times 12-3\times 2}&\frac{5}{5\times 12-3\times 2}\end{matrix}\right)\left(\begin{matrix}97\\109\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{2}{9}&-\frac{1}{18}\\-\frac{1}{27}&\frac{5}{54}\end{matrix}\right)\left(\begin{matrix}97\\109\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{9}\times 97-\frac{1}{18}\times 109\\-\frac{1}{27}\times 97+\frac{5}{54}\times 109\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{31}{2}\\\frac{13}{2}\end{matrix}\right)
Do the arithmetic.
x=\frac{31}{2},y=\frac{13}{2}
Extract the matrix elements x and y.
5x+3y=97,2x+12y=109
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.
2\times 5x+2\times 3y=2\times 97,5\times 2x+5\times 12y=5\times 109
To make 5x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 5.
10x+6y=194,10x+60y=545
Simplify.
10x-10x+6y-60y=194-545
Subtract 10x+60y=545 from 10x+6y=194 by subtracting like terms on each side of the equal sign.
6y-60y=194-545
Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.
-54y=194-545
Add 6y to -60y.
-54y=-351
Add 194 to -545.
y=\frac{13}{2}
Divide both sides by -54.
2x+12\times \frac{13}{2}=109
Substitute \frac{13}{2} for y in 2x+12y=109. Because the resulting equation contains only one variable, you can solve for x directly.
2x+78=109
Multiply 12 times \frac{13}{2}.
2x=31
Subtract 78 from both sides of the equation.
x=\frac{31}{2}
Divide both sides by 2.
x=\frac{31}{2},y=\frac{13}{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}