Solve for a, b
a=-497
b=503
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a+b=6,-a+b=1000
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.
a+b=6
Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.
a=-b+6
Subtract b from both sides of the equation.
-\left(-b+6\right)+b=1000
Substitute -b+6 for a in the other equation, -a+b=1000.
b-6+b=1000
Multiply -1 times -b+6.
2b-6=1000
Add b to b.
2b=1006
Add 6 to both sides of the equation.
b=503
Divide both sides by 2.
a=-503+6
Substitute 503 for b in a=-b+6. Because the resulting equation contains only one variable, you can solve for a directly.
a=-497
Add 6 to -503.
a=-497,b=503
The system is now solved.
a+b=6,-a+b=1000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\-1&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}6\\1000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\-1&1\end{matrix}\right))\left(\begin{matrix}1&1\\-1&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-1&1\end{matrix}\right))\left(\begin{matrix}6\\1000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-1&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-1&1\end{matrix}\right))\left(\begin{matrix}6\\1000\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-1&1\end{matrix}\right))\left(\begin{matrix}6\\1000\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-1\right)}&-\frac{1}{1-\left(-1\right)}\\-\frac{-1}{1-\left(-1\right)}&\frac{1}{1-\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}6\\1000\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}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}6\\1000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 6-\frac{1}{2}\times 1000\\\frac{1}{2}\times 6+\frac{1}{2}\times 1000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-497\\503\end{matrix}\right)
Do the arithmetic.
a=-497,b=503
Extract the matrix elements a and b.
a+b=6,-a+b=1000
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.
a+a+b-b=6-1000
Subtract -a+b=1000 from a+b=6 by subtracting like terms on each side of the equal sign.
a+a=6-1000
Add b to -b. Terms b and -b cancel out, leaving an equation with only one variable that can be solved.
2a=6-1000
Add a to a.
2a=-994
Add 6 to -1000.
a=-497
Divide both sides by 2.
-\left(-497\right)+b=1000
Substitute -497 for a in -a+b=1000. Because the resulting equation contains only one variable, you can solve for b directly.
497+b=1000
Multiply -1 times -497.
b=503
Subtract 497 from both sides of the equation.
a=-497,b=503
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}