Start with the basic equation y=mx+b{\displaystyle y=mx+b}. [2] X Expert Source Mario Banuelos, PhDAssistant Professor of Mathematics Expert Interview. 11 December 2021. If you do not know the equations, find them based on the information you have. Example: Your two lines are y=x+3{\displaystyle y=x+3} and y−12=−2x{\displaystyle y-12=-2x}. To get y{\displaystyle y} alone in the second equation, add 12 to each side: y=12−2x{\displaystyle y=12-2x}
For example, if you want to know where the lines y = x + 3 crosses y = 12 - 2x, you’d equate them by writing x + 3 = 12 - 2x. [3] X Expert Source Mario Banuelos, PhDAssistant Professor of Mathematics Expert Interview. 11 December 2021.
Example: x+3=12−2x{\displaystyle x+3=12-2x} Add 2x{\displaystyle 2x} to each side: 3x+3=12{\displaystyle 3x+3=12} Subtract 3 from each side: 3x=9{\displaystyle 3x=9} Divide each side by 3: x=3{\displaystyle x=3}.
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}
Example: x=3{\displaystyle x=3} and y=12−2x{\displaystyle y=12-2x} y=12−2(3){\displaystyle y=12-2(3)} y=12−6{\displaystyle y=12-6} y=6{\displaystyle y=6} This is the same answer as before. We did not make any mistakes.
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, they do not intersect. The x{\displaystyle x} terms will cancel out, and your equation will simplify to a false statement (such as 0=1{\displaystyle 0=1}). Write “the lines do not intersect” or no real solution" as your answer. If the two equations describe the same line, they “intersect” everywhere. The x{\displaystyle x} terms will cancel out and your equation will simplify to a true statement (such as 3=3{\displaystyle 3=3}). Write “the two lines are the same” as your answer.
Expand equations with parentheses to check whether they’re quadratics. For example, y=(x+3)(x){\displaystyle y=(x+3)(x)} is quadratic, since it expands into y=x2+3x. {\displaystyle y=x^{2}+3x. } Equations for a circle or ellipse have both an x2{\displaystyle x^{2}} and a y2{\displaystyle y^{2}} term. [8] X Research source [9] X Research source If you’re having trouble with these special cases, see the Tips section below.
Example: Find the intersection of x2+2x−y=−1{\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+1{\displaystyle y=x^{2}+2x+1} and y=x+7{\displaystyle y=x+7}. This example has one quadratic equation and one linear equation. Problems with two quadratic equations are solved in a similar way.
Example: y=x2+2x+1{\displaystyle y=x^{2}+2x+1} and y=x+7{\displaystyle y=x+7} x2+2x+1=x+7{\displaystyle x^{2}+2x+1=x+7}
Example: x2+2x+1=x+7{\displaystyle x^{2}+2x+1=x+7} Subtract x from each side: x2+x+1=7{\displaystyle x^{2}+x+1=7} Subtract 7 from each side: x2+x−6=0{\displaystyle x^{2}+x-6=0}
Example: x2+x−6=0{\displaystyle x^{2}+x-6=0} The goal of factoring is to find the two factors that multiply together to make this equation. Starting with the first term, we know x2{\displaystyle x^{2}} can divide into x, and x. Write down (x )(x ) = 0 to show this. The last term is -6. List each pair of factors that multiply to make negative six: −6∗1{\displaystyle -61}, −3∗2{\displaystyle -32}, −2∗3{\displaystyle -23}, and −1∗6{\displaystyle -16}. The middle term is x (which you could write as 1x). Add each pair of factors together until you get 1 as an answer. The correct pair of factors is −2∗3{\displaystyle -2*3}, since −2+3=1{\displaystyle -2+3=1}. Fill out the gaps in your answer with this pair of factors: (x−2)(x+3)=0{\displaystyle (x-2)(x+3)=0}.
Example (factoring): We ended up with the equation (x−2)(x+3)=0{\displaystyle (x-2)(x+3)=0}. If either of the factors in parentheses equal 0, the equation is true. One solution is x−2=0{\displaystyle x-2=0} → x=2{\displaystyle x=2}. The other solution is x+3=0{\displaystyle x+3=0} → x=−3{\displaystyle x=-3}. Example (quadratic equation or complete the square): If you used one of these methods to solve your equation, a square root will show up. For example, our equation becomes x=(−1+25)/2{\displaystyle x=(-1+{\sqrt {25}})/2}. Remember that a square root can simplify to two different solutions: 25=5∗5{\displaystyle {\sqrt {25}}=55}, and 25=(−5)∗(−5){\displaystyle {\sqrt {25}}=(-5)(-5)}. Write two equations, one for each possibility, and solve for x in each one.
One solution: The problems factor into two identical factors ((x-1)(x-1) = 0). When plugged into the quadratic formula, the square root term is 0{\displaystyle {\sqrt {0}}}. You only need to solve one equation. No real solution: There are no factors that satisfy the requirements (summing to the middle term). When plugged into the quadratic formula, you get a negative number under the square root sign (such as −2{\displaystyle {\sqrt {-2}}}). Write “no solution” as your answer.
Example: We found two solutions, x=2{\displaystyle x=2} and x=−3{\displaystyle x=-3}. One of our lines has the equation y=x+7{\displaystyle y=x+7}. Plug in y=2+7{\displaystyle y=2+7} and y=−3+7{\displaystyle y=-3+7}, then solve each equation to find that y=9{\displaystyle y=9} and y=4{\displaystyle y=4}.
Example: When we plugged in x=2{\displaystyle x=2}, we got y=9{\displaystyle y=9}, so one intersection is at (2, 9). The same process for our second solution tells us another intersection lies at (-3, 4).
