Show A System of those two equations can be solved (find where they intersect), either:
How to Solve using Algebra
An example will help:
Make both equations into "y=" format: They are both in "y=" format, so go straight to next step Set them equal to each other x2 - 5x + 7 = 2x + 1 Simplify into "= 0" format (like a standard Quadratic Equation) Subtract 2x from both sides: x2 - 7x + 7 = 1 Subtract 1 from both sides: x2 - 7x + 6 = 0 Solve the Quadratic Equation! (The hardest part for me) You can read how to solve Quadratic Equations, but here we will factor the Quadratic Equation: Start with: x2 - 7x + 6 = 0 Rewrite -7x as -x-6x: x2 - x - 6x + 6 = 0 Then: x(x-1) - 6(x-1) = 0 Then: (x-1)(x-6) = 0  Which gives us the solutions x=1 and x=6 Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers The matching y values are (also see Graph):
Our solution: the two points are (1,3) and (6,13) I think of it as three stages:
Combine into Quadratic Equation ⇒ Solve the Quadratic ⇒ Calculate the points SolutionsThere are three possible cases:
Time for another example!
Make both equations into "y=" format: First equation is: y - x2 = 7 - 5x Add x2 to both sides: y = x2 + 7 - 5x Second equation is: 4y - 8x = -21 Add 8x to both sides: 4y = 8x - 21 Divide all by 4: y = 2x - 5.25 Set them equal to each other x2 - 5x + 7 = 2x - 5.25 Simplify into "= 0" format (like a standard Quadratic Equation) Subtract 2x from both sides: x2 - 7x + 7 = -5.25 Add 5.25 to both sides: x2 - 7x + 12.25 = 0 Solve the Quadratic Equation! Using the Quadratic Formula from Quadratic Equations:
Just one solution! (The "discriminant" is 0) Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers The matching y value is:
Our solution: (3.5,1.75)
Kaboom! The cannon ball flies through the air, following a parabola: y = 2 + 0.12x - 0.002x2 The land slopes upward: y = 0.15x Where does the cannon ball land? Both equations are already in the "y =" format, so set them equal to each other: 0.15x = 2 + 0.12x - 0.002x2 Simplify into "= 0" format: Bring all terms to left: 0.002x2 + 0.15x - 0.12x - 2 = 0 Simplify: 0.002x2 + 0.03x - 2 = 0 Multiply by 500: x2 + 15x - 1000 = 0 Solve the Quadratic Equation: Split 15x into -25x+40x: x2 -25x + 40x - 1000 = 0 Then: x(x-25) + 40(x-25) = 0 Then: (x+40)(x-25) = 0 x = -40 or 25 The negative answer can be ignored, so x = 25 Use the linear equation to calculate matching "y" value: y = 0.15 x 25 = 3.75 So the cannonball impacts the slope at (25, 3.75) You can also find the answer graphically by using the Function Grapher: Both Variables SquaredSometimes BOTH terms of the quadratic can be squared:
The circle x2 + y2 = 25 And the straight line 3y - 2x = 6 First put the line in "y=" format: Move 2x to right hand side: 3y = 2x + 6 Divide by 3: y = 2x/3 + 2 NOW, Instead of making the circle into "y=" format, we can use substitution (replace "y" in the quadratic with the linear expression): Put y = 2x/3 + 2 into circle equation: x2 + (2x/3 + 2)2 = 25 Expand: x2 + 4x2/9 + 2(2x/3)(2) + 22 = 25 Multiply all by 9: 9x2 + 4x2 + 2(2x)(2)(3) + (9)(22) = (9)(25) Simplify: 13x2+ 24x + 36 = 225 Subtract 225 from both sides: 13x2+ 24x - 189 = 0 Now it is in standard Quadratic form, let's solve it: 13x2+ 24x - 189 = 0 Split 24x into 63x-39x: 13x2+ 63x - 39x - 189 = 0 Then: x(13x + 63) - 3(13x + 63) = 0 Then: (x - 3)(13x + 63) = 0 So: x = 3 or -63/13 Now work out y-values: Substitute x = 3 into linear equation:
Substitute x = -63/13 into linear equation:
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Learning Outcomes
A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form [latex]Ax+By+C=0[/latex]. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes. Intersection of a Parabola and a LineThere are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.
The graphs below illustrate possible solution sets for a system of equations involving a parabola and a line.
How To: Given a system of equations containing a line and a parabola, find the solution.
Solve the system of equations. [latex]\begin{array}{l}x-y=-1\hfill \\ y={x}^{2}+1\hfill \end{array}[/latex]
Yes, but because [latex]x[/latex] is squared in the second equation this could give us extraneous solutions for [latex]x[/latex]. For [latex]y=1[/latex]
[latex]\begin{align}&y={x}^{2}+1 \\ &y={x}^{2}+1 \\ &{x}^{2}=0 \\ &x=\pm \sqrt{0}=0 \end{align}[/latex] This gives us the same value as in the solution. For [latex]y=2[/latex] [latex]\begin{align}&y={x}^{2}+1 \\ &2={x}^{2}+1 \\ &{x}^{2}=1 \\ &x=\pm \sqrt{1}=\pm 1 \end{align}[/latex] Notice that [latex]-1[/latex] is an extraneous solution.
Solve the given system of equations by substitution. [latex]\begin{gathered}3x-y=-2 \\ 2{x}^{2}-y=0 \end{gathered}[/latex]
Show Solution Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
The graph below illustrates possible solution sets for a system of equations involving a circle and a line.
How To: Given a system of equations containing a line and a circle, find the solution.
Find the intersection of the given circle and the given line by substitution. [latex]\begin{gathered}{x}^{2}+{y}^{2}=5 \\ y=3x - 5 \end{gathered}[/latex]
Show Solution
Solve the system of nonlinear equations. [latex]\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ x - 3y=-10\hfill \end{array}[/latex] We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.
The figure below illustrates possible solution sets for a system of equations involving a circle and an ellipse.
Solve the system of nonlinear equations. [latex]\begin{align} {x}^{2}+{y}^{2}=26 \hspace{5mm} \left(1\right)\\ 3{x}^{2}+25{y}^{2}=100 \hspace{5mm} \left(2\right)\end{align}[/latex]
Find the solution set for the given system of nonlinear equations. [latex]\begin{gathered}4{x}^{2}+{y}^{2}=13\\ {x}^{2}+{y}^{2}=10\end{gathered}[/latex] In the following video, we present an example of how to solve a system of non-linear equations that represent the intersection of an ellipse and a hyperbola. Did you have an idea for improving this content? We’d love your input. |
Write two solutions of the equation x+y 13
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