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Finding the intersection algebraically refers to the process of determining the common solutions, or points of intersection, between two or more equations. This method can be employed to find the values of variables that satisfy all given equations simultaneously. Solving for intersections algebraically provides a powerful tool for analyzing and understanding the relationships between multiple mathematical expressions. By following a systematic set of rules and techniques, one can efficiently identify the points where multiple equations intersect in the coordinate plane, finding precise solutions to complex mathematical problems. In this article, we will explore various algebraic strategies and methods that can help us effectively find intersections between equations to solve real-world problems or further our understanding of mathematical concepts.
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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.
Steps
Find the intersection of two lines
- 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} and y−twelfth=−2x{displaystyle y-12=-2x} . In the second equation, to leave only y on the left side, you add 12 to both sides: y=twelfth−2x{displaystyle y=12-2x}
- Example: We know y=x+3{displaystyle y=x+3} and y=twelfth−2x{displaystyle y=12-2x} , therefore x+3=twelfth−2x{displaystyle x+3=12-2x} .
- For example: x+3=twelfth−2x{displaystyle x+3=12-2x}
- Add 2x{displaystyle 2x} on two sides:
- 3x+3=twelfth{displaystyle 3x+3=12}
- Subtract 3 from both sides:
- 3x=9{displaystyle 3x=9}
- Divide both sides by 3:
- x=3{displaystyle x=3} .
- For example: x=3{displaystyle x=3} and y=x+3{displaystyle y=x+3}
- y=3+3{displaystyle y=3+3}
- y=6{displaystyle y=6}
- For example: x=3{displaystyle x=3} and y=twelfth−2x{displaystyle y=12-2x}
- y=twelfth−2(3){displaystyle y=12-2(3)}
- y=twelfth−6{displaystyle y=12-6}
- y=6{displaystyle y=6}
- Thus we get the same y value. The essay has no errors.
- For example: x=3{displaystyle x=3} and y=6{displaystyle y=6}
- The two lines intersect at (3,6).
- 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} ). 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} ). Write the answer as ” two lines coincide “.
Problems with quadratic equations
- Expand the equations from parentheses to check if they have quadratic form. For example, y=(x+3)(x){displaystyle y=(x+3)(x)} has quadratic form because it is expanded to y=x2+3x.{displaystyle y=x^{2}+3x.}
- Equations of circles and ellipses have both terms x2{displaystyle x^{2}} and y2{displaystyle y^{2}} . [1] X Research Sources[2] X Research Resources If you are having trouble with these special cases see the Advice below.
- Example: Find the intersection of x2+2x−y=−first{displaystyle x^{2}+2x-y=-1} and y=x+7{displaystyle y=x+7} .
- Rewrite the quadratic equation in terms of y:
- y=x2+2x+first{displaystyle y=x^{2}+2x+1} and y=x+7{displaystyle y=x+7} .
- This example has a quadratic equation and a linear equation. Problems with two quadratic equations are solved similarly.
- For example: y=x2+2x+first{displaystyle y=x^{2}+2x+1} and y=x+7{displaystyle y=x+7}
- x2+2x+first=x+7{displaystyle x^{2}+2x+1=x+7}
- For example:x2+2x+first=x+7{displaystyle x^{2}+2x+1=x+7}
- Subtract x from both sides:
- x2+x+first=7{displaystyle x^{2}+x+1=7}
- Subtract 7 from both sides:
- x2+x−6=0{displaystyle x^{2}+x-6=0}
- For example: x2+x−6=0{displaystyle 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 x2{displaystyle 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} , −3∗2{displaystyle -3*2} , −2∗3{displaystyle -2*3} , and −first∗6{displaystyle -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} , because −2+3=first{displaystyle -2+3=1} .
- Fill in the blanks with this pair of factors: (x−2)(x+3)=0{displaystyle (x-2)(x+3)=0} .
- Example (factoring): Finally we have the equation (x−2)(x+3)=0{displaystyle (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{displaystyle x=2} . The remaining solution is x+3=0{displaystyle x+3=0} → x=−3{displaystyle 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} . Remember that square roots can be simplified to two different solutions: 25=5∗5{displaystyle {sqrt {25}}=5*5} , and 25=(−5)∗(−5){displaystyle {sqrt {25}}=(-5)*(-5)} . Write two equations for each case and solve for x respectively.
- 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}}} . 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}}} ). Write the answer as “no solution”.
- Example: We find two solutions x=2{displaystyle x=2} and x=−3{displaystyle x=-3} . One of the two lines has the equation y=x+7{displaystyle y=x+7} . Instead y=2+7{displaystyle y=2+7} and y=−3+7{displaystyle y=-3+7} , then solve each equation to find y=9{displaystyle y=9} and y=4{displaystyle y=4} .
- Example: When replacing x=2{displaystyle x=2} in, we have y=9{displaystyle y=9} , so the intersection has coordinates (2, 9) . Doing the same for the second solution will give the coordinates of the remaining intersection as (-3, 4) .
Advice
- The equations of a circle and an ellipse have a term x2{displaystyle x^{2}} and a number of grades y2{displaystyle 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 x2{displaystyle x^{2}} and replace the answer in x2{displaystyle 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.
wikiHow is a “wiki” site, which means that many of the articles here are written by multiple authors. To create this article, 16 people, some of whom are anonymous, have edited and improved the article over time.
This article has been viewed 63,431 times.
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.
In conclusion, finding the intersection algebraically is a simple and effective method for determining the common solutions of two equations. By setting the two equations equal to each other and solving for the unknown variable, we can identify the numerical value(s) at which the two equations intersect. This method is particularly helpful in solving simultaneous equations, as it provides a clear and concise solution. Additionally, algebraic methods allow for a deeper understanding of the relationship between the two equations and can be extended to solve equations with more variables. Overall, algebraic methods provide a powerful tool for finding intersections and offer a valuable approach in various mathematical applications.
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