Solve for x, y
x=\frac{p-1}{2}
y=-p
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y+p=0,y+2x+1=0
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.
y+p=0
Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.
y=-p
Subtract p from both sides of the equation.
-p+2x+1=0
Substitute -p for y in the other equation, y+2x+1=0.
2x+1-p=0
Add -p to 1.
2x=p-1
Subtract -p+1 from both sides of the equation.
x=\frac{p-1}{2}
Divide both sides by 2.
y=-p,x=\frac{p-1}{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}