How to Find Intersection Algebraically

When two lines intersect on a two-dimensional coordinate system, they meet only at a point represented by the x and y coordinate pair. Since both lines pass through that point, the x, y coordinate pair must satisfy both equations. With some additional techniques, you can find the intersection of the parabola and other quadratic curves by analogous reasoning.

Table of Contents

Find the intersection of two lines

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Write the equation of each line with y being on the left side. If necessary, transform the equation so that only y is on one side of the equal sign. If the equation uses f(x) or g(x) instead of y, separate this term. Remember that you can cancel terms by performing the same operation on both sides.

If the problem doesn’t say equations, find them from the information you already have.
Example: Two lines whose equation is $y$ $=$ $x$ $+$ $3$ ${displaystyle y=x+3}$ $y=x+3$ and $y$ $-$ $twelfth$ $=$ $-$ $2$ $x$ ${displaystyle y-12=-2x}$ $y-12=-2x$ . In the second equation, to leave only y on the left side, you add 12 to both sides: $y$ $=$ $twelfth$ $-$ $2$ $x$ ${displaystyle y=12-2x}$ $y=12-2x$

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Set the right side of the two equations equal. We are looking for the point where two lines have the same x, y coordinates; This is where the two lines intersect. Both equations have only y on the left side, so their right sides will be equal. Write a new equation to represent this.

Example: We know $y$ $=$ $x$ $+$ $3$ ${displaystyle y=x+3}$ $y=x+3$ and $y$ $=$ $twelfth$ $-$ $2$ $x$ ${displaystyle y=12-2x}$ $y=12-2x$ , therefore $x$ $+$ $3$ $=$ $twelfth$ $-$ $2$ $x$ ${displaystyle x+3=12-2x}$ $x+3=12-2x$ .

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Solve for x. The new equation has only one variable x. Solve the equation algebraically, that is, do the same math on both sides. Convert all terms with x to one side of the equation, then return it to the form x = __. (If you can’t, scroll down to the bottom of this section.)

For example: $x$ $+$ $3$ $=$ $twelfth$ $-$ $2$ $x$ ${displaystyle x+3=12-2x}$ $x+3=12-2x$
Add $2$ $x$ ${displaystyle 2x}$ $2x$ on two sides:
$3$ $x$ $+$ $3$ $=$ $twelfth$ ${displaystyle 3x+3=12}$ $3x+3=12$
Subtract 3 from both sides:
$3$ $x$ $=$ $9$ ${displaystyle 3x=9}$ $3x=9$
Divide both sides by 3:
$x$ $=$ $3$ ${displaystyle x=3}$ $x=3$ .

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Use the x value to find y. Choose the equation of either line. Substitute the x value you just found into this equation. Solve for y by arithmetic method.

For example: $x$ $=$ $3$ ${displaystyle x=3}$ $x=3$ and $y$ $=$ $x$ $+$ $3$ ${displaystyle y=x+3}$ $y=x+3$
$y$ $=$ $3$ $+$ $3$ ${displaystyle y=3+3}$ $y=3+3$
$y$ $=$ $6$ ${displaystyle y=6}$ $y=6$

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Check the result. You should substitute the value of x in the other equation to see if you get the same result. If you get a different y value, you must check your work.

For example: $x$ $=$ $3$ ${displaystyle x=3}$ $x=3$ and $y$ $=$ $twelfth$ $-$ $2$ $x$ ${displaystyle y=12-2x}$ $y=12-2x$
$y$ $=$ $twelfth$ $-$ $2$ $($ $3$ $)$ ${displaystyle y=12-2(3)}$ $y=12-2(3)$
$y$ $=$ $twelfth$ $-$ $6$ ${displaystyle y=12-6}$ $y=12-6$
$y$ $=$ $6$ ${displaystyle y=6}$ $y=6$
Thus we get the same y value. The essay has no errors.

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Write a pair of x,y coordinates of the intersection. You have now found the x, y coordinate pair where the two lines intersect. Write out this point as a pair of coordinates, with the x value coming first.

For example: $x$ $=$ $3$ ${displaystyle x=3}$ $x=3$ and $y$ $=$ $6$ ${displaystyle y=6}$ $y=6$
The two lines intersect at (3,6).

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Handling unusual cases. Some equations cannot be solved to find x. This is not necessarily because you made a mistake. The equation of a pair of lines can have unusual solutions in the following two cases:

If the two lines are parallel, then they do not intersect. The x terms are canceled out and the equation is simplified to a false statement (e.g. $0$ $=$ $first$ ${displaystyle 0=1}$ $0=1$ ). Write the answer as ” two lines do not intersect ” or ” no real solution “.
If two equations represent the same line, they “intersect” at every point. The x terms are canceled out and the equation is simplified to a true statement (e.g. $3$ $=$ $3$ ${displaystyle 3=3}$ $3=3$ ). Write the answer as ” two lines coincide “.

Problems with quadratic equations

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Identify quadratic equations. In a quadratic equation, one or more variables will have powers (

x

2

{displaystyle x^{2}}

x^{2}

or

y

2

{displaystyle y^{2}}

y^{2}

), and no variable has a higher power. The graphs of these equations are curvilinear, so they can intersect the line at 0, 1, or 2 points. This section shows you how to find those intersections of the problem.

Expand the equations from parentheses to check if they have quadratic form. For example, $y$ $=$ $($ $x$ $+$ $3$ $)$ $($ $x$ $)$ ${displaystyle y=(x+3)(x)}$ $y=(x+3)(x)$ has quadratic form because it is expanded to $y$ $=$ $x$ $2$ $+$ $3$ $x$ $.$ ${displaystyle y=x^{2}+3x.}$ $y=x^{2}+3x.$
Equations of circles and ellipses have both terms $x$ $2$ ${displaystyle x^{2}}$ $x^{2}$ and $y$ $2$ ${displaystyle y^{2}}$ $y^{2}$ . ^{[1] X Research Sources}^{[2] X Research Resources} If you are having trouble with these special cases see the Advice below.

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Write the equations in terms of y. If necessary, transform each side of the equation so that only y is on one side of the equal sign.

Example: Find the intersection of $x$ $2$ $+$ $2$ $x$ $-$ $y$ $=$ $-$ $first$ ${displaystyle x^{2}+2x-y=-1}$ $x^{2}+2x-y=-1$ and $y$ $=$ $x$ $+$ $7$ ${displaystyle y=x+7}$ $y=x+7$ .
Rewrite the quadratic equation in terms of y:
$y$ $=$ $x$ $2$ $+$ $2$ $x$ $+$ $first$ ${displaystyle y=x^{2}+2x+1}$ $y=x^{2}+2x+1$ and $y$ $=$ $x$ $+$ $7$ ${displaystyle y=x+7}$ $y=x+7$ .
This example has a quadratic equation and a linear equation. Problems with two quadratic equations are solved similarly.

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Combine the two equations to cancel y. After you convert both sides of the equations in terms of y, the two sides without y will be equal.

For example: $y$ $=$ $x$ $2$ $+$ $2$ $x$ $+$ $first$ ${displaystyle y=x^{2}+2x+1}$ $y=x^{2}+2x+1$ and $y$ $=$ $x$ $+$ $7$ ${displaystyle y=x+7}$ $y=x+7$
$x$ $2$ $+$ $2$ $x$ $+$ $first$ $=$ $x$ $+$ $7$ ${displaystyle x^{2}+2x+1=x+7}$ $x^{2}+2x+1=x+7$

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Transform the new equation so that one side is zero. Use algebra to convert all terms to one side. The problem is then ready to be solved in the next step.

For example: $x$ $2$ $+$ $2$ $x$ $+$ $first$ $=$ $x$ $+$ $7$ ${displaystyle x^{2}+2x+1=x+7}$ $x^{2}+2x+1=x+7$
Subtract x from both sides:
$x$ $2$ $+$ $x$ $+$ $first$ $=$ $7$ ${displaystyle x^{2}+x+1=7}$ $x^{2}+x+1=7$
Subtract 7 from both sides:
$x$ $2$ $+$ $x$ $-$ $6$ $=$ $0$ ${displaystyle x^{2}+x-6=0}$ $x^{2}+x-6=0$

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Solve quadratic equations . After turning the equation to zero, you have three solutions, the choice of which solution will be up to each person. You can learn how to use the square root formula or the “squad complement” method, or see the example below for factoring:

For example: $x$ $2$ $+$ $x$ $-$ $6$ $=$ $0$ ${displaystyle x^{2}+x-6=0}$ $x^{2}+x-6=0$
The purpose of factoring is to find two factors that, when multiplied together, form an equation. Starting with the first term, we know $x$ $2$ ${displaystyle x^{2}}$ $x^{2}$ can be decomposed into x and x. Write it in the form (x )(x ) = 0.
The final term is -6. List each pair of factors that, when multiplied together, equal -6: $-$ $6$ $*$ $first$ ${displaystyle -6*1}$ $-6*1$ , $-$ $3$ $*$ $2$ ${displaystyle -3*2}$ $-3*2$ , $-$ $2$ $*$ $3$ ${displaystyle -2*3}$ $-2*3$ , and $-$ $first$ $*$ $6$ ${displaystyle -1*6}$ $-1*6$ .
The middle term is x (which can be written as 1x). Add each pair of factors together until you get 1. The correct pair of factors is $-$ $2$ $*$ $3$ ${displaystyle -2*3}$ $-2*3$ , because $-$ $2$ $+$ $3$ $=$ $first$ ${displaystyle -2+3=1}$ $-2+3=1$ .
Fill in the blanks with this pair of factors: $($ $x$ $-$ $2$ $)$ $($ $x$ $+$ $3$ $)$ $=$ $0$ ${displaystyle (x-2)(x+3)=0}$ $(x-2)(x+3)=0$ .

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Notice that we have two solutions x. If you solve it too quickly, you may find only one solution without realizing there is a second solution. Here’s how to find two solutions x for lines that intersect at two points:

Example (factoring): Finally we have the equation $($ $x$ $-$ $2$ $)$ $($ $x$ $+$ $3$ $)$ $=$ $0$ ${displaystyle (x-2)(x+3)=0}$ $(x-2)(x+3)=0$ . If either factor is 0, then the equation is satisfied. One solution is $x$ $-$ $2$ $=$ $0$ ${displaystyle x-2=0}$ $x-2=0$ → $x$ $=$ $2$ ${displaystyle x=2}$ $x=2$ . The remaining solution is $x$ $+$ $3$ $=$ $0$ ${displaystyle x+3=0}$ $x+3=0$ → $x$ $=$ $-$ $3$ ${displaystyle x=-3}$ $x=-3$ .
Example (quadratic solution formula or square’s complement): If you use either of these two ways to solve the equation, the square root sign will appear. For example, the equation becomes $x$ $=$ $($ $-$ $first$ $+$ $25$ $)$ $/$ $2$ ${displaystyle x=(-1+{sqrt {25}})/2}$ $x=(-1+{sqrt {25}})/2$ . Remember that square roots can be simplified to two different solutions: $25$ $=$ $5$ $*$ $5$ ${displaystyle {sqrt {25}}=5*5}$ ${sqrt {25}}=5*5$ , and $25$ $=$ $($ $-$ $5$ $)$ $*$ $($ $-$ $5$ $)$ ${displaystyle {sqrt {25}}=(-5)*(-5)}$ ${sqrt {25}}=(-5)*(-5)$ . Write two equations for each case and solve for x respectively.

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Solve problems with one or no solutions. Two lines that touch each other only have one intersection, and two lines that never touch each other have no intersection. Here’s how to know:

One solution: The problem can be factored into two identical factors ((x-1)(x-1) = 0). When substituting in the quadratic formula, the term whose root is $0$ ${displaystyle {sqrt {0}}}$ ${sqrt {0}}$ . You only need to solve one equation.
No real solution: No factor can satisfy the requirement (sum equals the middle term). When you substitute in the square root formula, you get a negative number below the square root sign (for example, $-$ $2$ ${displaystyle {sqrt {-2}}}$ ${sqrt {-2}}$ ). Write the answer as “no solution”.

Advice

The equations of a circle and an ellipse have a term $x$ $2$ ${displaystyle x^{2}}$ $x^{2}$ and a number of grades $y$ $2$ ${displaystyle y^{2}}$ $y^{2}$ . To find the intersection of the circle and the line, solve for x in the linear equation. Substitute the solution for x in the equation of the circle and you will have a quadratic equation that is easier to solve. These problems can have 0, 1, or 2 solutions, as described in the method above.
The circle and the parabp (or other quadratic) can have 0, 1, 2, 3, or 4 solutions. Find the variable to the power of 2 in both equations — say x ² . Solve find $x$ $2$ ${displaystyle x^{2}}$ $x^{2}$ and replace the answer in $x$ $2$ ${displaystyle x^{2}}$ $x^{2}$ in the remaining equation. Solve for y to get 0, 1 or 2 solutions. Substitute each solution back into the original quadratic equation to solve for x. Each of these equations can have 0, 1, or 2 solutions.

Steps

Find the intersection of two lines

Problems with quadratic equations

Advice

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