\left\{ \begin{array} { l } { x + y = 1 } \\ { x ^ { 2 } + y ^ { 2 } = 3 } \end{array} \right.
Solve for x, y
x=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }y=\frac{1-\sqrt{5}}{2}\approx -0.618033989
x=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }y=\frac{\sqrt{5}+1}{2}\approx 1.618033989
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x+y=1,y^{2}+x^{2}=3
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=1
Solve x+y=1 for x by isolating x on the left hand side of the equal sign.
x=-y+1
Subtract y from both sides of the equation.
y^{2}+\left(-y+1\right)^{2}=3
Substitute -y+1 for x in the other equation, y^{2}+x^{2}=3.
y^{2}+y^{2}-2y+1=3
Square -y+1.
2y^{2}-2y+1=3
Add y^{2} to y^{2}.
2y^{2}-2y-2=0
Subtract 3 from both sides of the equation.
y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-2\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 1\left(-1\right)\times 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-2\right)}}{2\times 2}
Square 1\times 1\left(-1\right)\times 2.
y=\frac{-\left(-2\right)±\sqrt{4-8\left(-2\right)}}{2\times 2}
Multiply -4 times 1+1\left(-1\right)^{2}.
y=\frac{-\left(-2\right)±\sqrt{4+16}}{2\times 2}
Multiply -8 times -2.
y=\frac{-\left(-2\right)±\sqrt{20}}{2\times 2}
Add 4 to 16.
y=\frac{-\left(-2\right)±2\sqrt{5}}{2\times 2}
Take the square root of 20.
y=\frac{2±2\sqrt{5}}{2\times 2}
The opposite of 1\times 1\left(-1\right)\times 2 is 2.
y=\frac{2±2\sqrt{5}}{4}
Multiply 2 times 1+1\left(-1\right)^{2}.
y=\frac{2\sqrt{5}+2}{4}
Now solve the equation y=\frac{2±2\sqrt{5}}{4} when ± is plus. Add 2 to 2\sqrt{5}.
y=\frac{\sqrt{5}+1}{2}
Divide 2+2\sqrt{5} by 4.
y=\frac{2-2\sqrt{5}}{4}
Now solve the equation y=\frac{2±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from 2.
y=\frac{1-\sqrt{5}}{2}
Divide 2-2\sqrt{5} by 4.
x=-\frac{\sqrt{5}+1}{2}+1
There are two solutions for y: \frac{1+\sqrt{5}}{2} and \frac{1-\sqrt{5}}{2}. Substitute \frac{1+\sqrt{5}}{2} for y in the equation x=-y+1 to find the corresponding solution for x that satisfies both equations.
x=-\frac{1-\sqrt{5}}{2}+1
Now substitute \frac{1-\sqrt{5}}{2} for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.
x=-\frac{\sqrt{5}+1}{2}+1,y=\frac{\sqrt{5}+1}{2}\text{ or }x=-\frac{1-\sqrt{5}}{2}+1,y=\frac{1-\sqrt{5}}{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}