\left\{ \begin{array} { l } { y = \frac { 1 } { 2 } x - 1 } \\ { y = - \frac { \sqrt { 5 } } { 2 } x + \sqrt { 5 } } \end{array} \right.
Solve for y, x
x=2
y=0
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y-\frac{1}{2}x=-1
Consider the first equation. Subtract \frac{1}{2}x from both sides.
y=\frac{-\sqrt{5}x}{2}+\sqrt{5}
Consider the second equation. Express \left(-\frac{\sqrt{5}}{2}\right)x as a single fraction.
y-\frac{-\sqrt{5}x}{2}=\sqrt{5}
Subtract \frac{-\sqrt{5}x}{2} from both sides.
2y+\sqrt{5}x=2\sqrt{5}
Multiply both sides of the equation by 2.
y-\frac{1}{2}x=-1,2y+\sqrt{5}x=2\sqrt{5}
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-\frac{1}{2}x=-1
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
y=\frac{1}{2}x-1
Add \frac{x}{2} to both sides of the equation.
2\left(\frac{1}{2}x-1\right)+\sqrt{5}x=2\sqrt{5}
Substitute \frac{x}{2}-1 for y in the other equation, 2y+\sqrt{5}x=2\sqrt{5}.
x-2+\sqrt{5}x=2\sqrt{5}
Multiply 2 times \frac{x}{2}-1.
\left(\sqrt{5}+1\right)x-2=2\sqrt{5}
Add x to \sqrt{5}x.
\left(\sqrt{5}+1\right)x=2\sqrt{5}+2
Add 2 to both sides of the equation.
x=2
Divide both sides by 1+\sqrt{5}.
y=\frac{1}{2}\times 2-1
Substitute 2 for x in y=\frac{1}{2}x-1. Because the resulting equation contains only one variable, you can solve for y directly.
y=1-1
Multiply \frac{1}{2} times 2.
y=0
Add -1 to 1.
y=0,x=2
The system is now solved.
y-\frac{1}{2}x=-1
Consider the first equation. Subtract \frac{1}{2}x from both sides.
y=\frac{-\sqrt{5}x}{2}+\sqrt{5}
Consider the second equation. Express \left(-\frac{\sqrt{5}}{2}\right)x as a single fraction.
y-\frac{-\sqrt{5}x}{2}=\sqrt{5}
Subtract \frac{-\sqrt{5}x}{2} from both sides.
2y+\sqrt{5}x=2\sqrt{5}
Multiply both sides of the equation by 2.
y-\frac{1}{2}x=-1,2y+\sqrt{5}x=2\sqrt{5}
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.
2y+2\left(-\frac{1}{2}\right)x=2\left(-1\right),2y+\sqrt{5}x=2\sqrt{5}
To make y and 2y equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.
2y-x=-2,2y+\sqrt{5}x=2\sqrt{5}
Simplify.
2y-2y-x+\left(-\sqrt{5}\right)x=-2-2\sqrt{5}
Subtract 2y+\sqrt{5}x=2\sqrt{5} from 2y-x=-2 by subtracting like terms on each side of the equal sign.
-x+\left(-\sqrt{5}\right)x=-2-2\sqrt{5}
Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.
\left(-\sqrt{5}-1\right)x=-2-2\sqrt{5}
Add -x to -\sqrt{5}x.
\left(-\sqrt{5}-1\right)x=-2\sqrt{5}-2
Add -2 to -2\sqrt{5}.
x=2
Divide both sides by -1-\sqrt{5}.
2y+\sqrt{5}\times 2=2\sqrt{5}
Substitute 2 for x in 2y+\sqrt{5}x=2\sqrt{5}. Because the resulting equation contains only one variable, you can solve for y directly.
2y+2\sqrt{5}=2\sqrt{5}
Multiply \sqrt{5} times 2.
2y=0
Subtract 2\sqrt{5} from both sides of the equation.
y=0
Divide both sides by 2.
y=0,x=2
The system is now solved.
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Simultaneous equation
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Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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