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Lesson Explainer: Solving Systems of Equations Graphically Mathematics

In this explainer, we will learn how to solve a system of two linear equations or one linear and one quadratic equation by considering their graphs and identifying the point of intersection.

When we graph any function, say 𝑦=𝑓(π‘₯), the π‘₯-coordinate of any point on the graph tells us the input value of the function and the 𝑦-coordinate tells us the corresponding output. In other words, every point on the graph is of the form (π‘₯,𝑓(π‘₯)). In the same way, if we sketch the graph of an equation, then the coordinates of any point on the graph satisfy the equation.

We can use this idea to solve systems of equations by using their graphs. If we sketch the graphs of two equations on the same pair of coordinate axes, then any point that lies on both graphs of the equations will satisfy both equations; it will be a solution to the system.

For example, consider the following system of linear equations: 𝑦=3π‘₯+2,𝑦=βˆ’4π‘₯+16.

We can solve this system by sketching both lines on the same pair of coordinate axes and finding the coordinates of the point of intersection. There are multiple ways for us to sketch a straight line; for instance, one way is to find the coordinates of two points on it and connect them together, continuing the line onward on either side. We substitute π‘₯=0 and π‘₯=1 into each equation to find two points on each line: 𝑦=3(0)+2=2,𝑦=3(1)+2=5,𝑦=βˆ’4(0)+16=16,𝑦=βˆ’4(1)+16=12.

Thus, the first line connects the points (0,2) and (1,5) and the second line connects the points (0,16) and (1,12). We can sketch these lines as follows.

We see that there is a single point of intersection between the lines with coordinates (2,8). Since it lies on the graphs of both equations, π‘₯=2 and 𝑦=8 must be solutions to both equations. Hence, this is a solution to the system of equations.

It is worth noting that every solution to a system of equations is a point of intersection on the graphs. This means that since there is only one point of intersection, there is only one solution to the system.

In our first example, we will identify the solution to a system of equations by using their graphs.

Example 1: Identifying the Solutions to a System of Equations from Their Graphs

Use the shown graph to solve the given simultaneous equations 𝑦=4π‘₯βˆ’2,𝑦=βˆ’π‘₯+3.

Answer

When we graph a function, the π‘₯- and 𝑦-coordinates of any point on the graph of the equation tell us values of π‘₯ and 𝑦 that solve that equation. Therefore, any point that lies on both graphs of the equations will satisfy both equations; it will be a solution to the system. We can see that there is only one point of intersection, and we can read the coordinates from the graph.

We see that the point of intersection is (1,2). Thus, π‘₯=1 and 𝑦=2 is a solution to both equations (since the point lies on both lines).

Hence, π‘₯=1 and 𝑦=2 is the solution to the system of equations.

In our next example, we will find the solution to a system of linear equations by sketching graphs of both equations and finding the point of intersection.

Example 2: Drawing a Linear System of Equations and Solving Them

By plotting the graphs of 𝑦=π‘₯βˆ’1 and 𝑦=5π‘₯+7, find the point that satisfies both equations simultaneously.

Answer

If a specific pair of values of π‘₯ and 𝑦 satisfies both equations, then we can note that the point with these coordinates must lie on the intersection between both graphs. This is because we plot graphs by plotting the values of π‘₯ and 𝑦 that satisfy the equation, so the point of intersection satisfies both equations.

There are many ways of sketching the graphs of straight lines. For example, we could use the fact that both lines are given in the slope–intercept form 𝑦=π‘šπ‘₯+𝑐. So, the slope of each line is π‘š and the 𝑦-intercept is 𝑐.

Alternatively, we can sketch any straight line by finding the coordinates of two points on the line and then connecting these points with a straight line. We note that the 𝑦-intercept of 𝑦=π‘₯βˆ’1 is (0,βˆ’1), and we can substitute π‘₯=1 into the equation to get 𝑦=1βˆ’1=0.

Thus, the graph of 𝑦=π‘₯βˆ’1 passes through (0,βˆ’1) and (1,0), so we can connect these points with a straight line to sketch its graph.

We follow the same process for 𝑦=5π‘₯+7. We note that its 𝑦-intercept is at (0,7), and we substitute π‘₯=1 into the equation to get 𝑦=5(1)+7=12.

Thus, the graph of 𝑦=5π‘₯+7 passes through (0,7) and (1,12), and we can connect these points with a straight line to sketch its graph.

In our sketch, we can see that the point of intersection between the lines has coordinates (βˆ’2,βˆ’3). Since this is only a sketch, it is a good idea to verify that these values satisfy both equations. We can do this by substituting π‘₯=βˆ’2 and 𝑦=βˆ’3 into both equations.

For the first equation, we have 𝑦=π‘₯βˆ’1βˆ’3=βˆ’2βˆ’1=βˆ’3, and for the second equation, we have 𝑦=5π‘₯+7βˆ’3=5(βˆ’2)+7=βˆ’3.

Thus, π‘₯=βˆ’2 and 𝑦=βˆ’3 satisfy both equations. Hence, (βˆ’2,βˆ’3) is the point of intersection between the lines.

We can apply this process as long as we can graph the equations. In our next example, we will identify the solutions to a linear–quadratic system of equations by using their given graphs.

Example 3: Identifying the Solutions to a System of Equations from Their Graphs

Use the shown graph to solve the simultaneous equations 𝑦=βˆ’2π‘₯+1,𝑦=βˆ’π‘₯βˆ’2π‘₯+1.

Answer

We are given the graphs of both equations and asked to use these graphs to solve the simultaneous equations. We can do this by recalling that the solution to a system of simultaneous equations is given by the coordinates of the point of intersection between the graphs of the equations.

In the given graph, we can see that there is a single point of intersection at (0,1).

Since every solution to the system of equations is given by the points of intersection, we can conclude that this is the only solution. We can verify this solution by substituting π‘₯=0 and 𝑦=1 into both equations to check that they satisfy each equation.

Hence, the only solution to the system is π‘₯=0 and 𝑦=1.

In the previous example, we saw that a given linear–quadratic system of equations only had a single solution. We can prove this by recalling that the sign of the discriminant of a quadratic tells us the number of roots of a quadratic function.

We can solve the system of equations algebraically by setting the two expressions for 𝑦 to be equal: βˆ’2π‘₯+1=βˆ’π‘₯βˆ’2π‘₯+1.

Then, we can rearrange and collect like terms: βˆ’2π‘₯+1+π‘₯+2π‘₯βˆ’1=βˆ’π‘₯βˆ’2π‘₯+1+π‘₯+2π‘₯βˆ’1π‘₯+(βˆ’2+2)π‘₯+(1βˆ’1)=0.

Following this, we can determine the number of solutions to a quadratic equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 by considering the sign of the discriminant π‘βˆ’4π‘Žπ‘οŠ¨. We have π‘Ž=1, 𝑏=0, and 𝑐=0, so π‘βˆ’4π‘Žπ‘=(0)βˆ’4(1)(0)=0.

Since the discriminant is equal to 0, we can conclude that there is only 1 root of the quadratic. Hence, there is only one solution to the system of equations.

In general, if we have a linear–quadratic system of equations of the form 𝑦=𝑝π‘₯+π‘ž,𝑦=π‘Žπ‘₯+𝑠π‘₯+𝑑, then we can set the expressions for 𝑦 to be equal to each other to get 𝑝π‘₯+π‘ž=π‘Žπ‘₯+𝑠π‘₯+𝑑.

We can then rearrange to obtain π‘Žπ‘₯+(π‘ βˆ’π‘)π‘₯+(π‘‘βˆ’π‘ž)=0.

If we relabel the coefficients, we can see that we have π‘Žπ‘₯+𝑏π‘₯+𝑐=0.

Assuming π‘Ž is nonzero, we can then check the sign of the discriminant to determine the number of roots of this equation and, hence, the number of solutions to the system of equations:

  • If π‘βˆ’4π‘Žπ‘>0, then the system has 2 solutions.
  • If π‘βˆ’4π‘Žπ‘=0, then the system has 1 solution.
  • If π‘βˆ’4π‘Žπ‘<0, then the system has no real solutions.

We can see each of these cases by considering how we can sketch quadratics and lines on the plane.

There are two points of intersection, π‘βˆ’4π‘Žπ‘>0; so, there are two real solutions.

There is one point of intersection, π‘βˆ’4π‘Žπ‘=0; so, there is one real solution.

There are no points of intersection, π‘βˆ’4π‘Žπ‘<0; so, there are no real solutions.

This gives us the following property of systems of equations.

Property: Number of Solutions to a Linear–Quadratic System

  • If we can rearrange a linear–quadratic system of equations into the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Žβ‰ 0, then
    • if π‘βˆ’4π‘Žπ‘>0, then the system has 2 solutions;
    • if π‘βˆ’4π‘Žπ‘=0, then the system has 1 solution;
    • if π‘βˆ’4π‘Žπ‘<0, then the system has no real solutions.

In our next example, we will identify the solutions to a linear–quadratic system of equations by using their given graphs.

Example 4: Identifying the Solutions to a System of Equations from Their Graphs

Use the shown graph to find the solutions of the simultaneous equations 𝑦=3π‘₯βˆ’1,𝑦=3π‘₯+2π‘₯+1.

Answer

We are given the graphs of both equations and asked to use these graphs to solve the simultaneous equations. We can do this by recalling that the solutions to a system of simultaneous equations are given by the coordinates of the point of intersection between the graphs of the equations.

We can see from the given diagram that the graphs of the two equations do not intersect. We can conclude that since there is no point of intersection, the systems of equations have no solutions.

It is worth noting that we can show why this is the case by considering an algebraic method of solving the system. We set the expressions for 𝑦 to be equal to each other to get 3π‘₯βˆ’1=3π‘₯+2π‘₯+1.

We subtract 3π‘₯βˆ’1 from both sides of the equation to obtain 0=3π‘₯βˆ’π‘₯+2.

We can then determine the number of solutions to the system by finding the sign of the discriminant ο€Ήπ‘βˆ’4π‘Žπ‘ο…οŠ¨ of this quadratic. We have (βˆ’1)βˆ’4(3)(2)=βˆ’23.

Since the discriminant is negative, we can conclude that the quadratic and the system have no real solutions.

In our next example, we will find the solutions to a linear–quadratic system of equations by sketching the graph of each equation and finding the coordinates of the points of intersection.

Example 5: Finding the Set of Points of the Intersection of Two Graphs

Find the set of points of intersection of the graphs of π‘₯+𝑦=8 and π‘₯+𝑦=50.

Answer

We first recall that the points of intersection between the two graphs will be given by the solution to the system of both equations: π‘₯+𝑦=8,π‘₯+𝑦=50.

We can solve this system in two ways. First, we can recall that the points of intersection of their graphs will give the solutions to the system. We can graph both equations using a graphing software.

We can actually note that the equation π‘₯+𝑦=50 gives a circle of radius √50 centered at the origin; we see that there are two points of intersection: (1,7) and (7,1).

We can also solve the system algebraically to find the points of intersection. We first find an expression for 𝑦 by subtracting π‘₯ from both sides of the first equation. This gives us 𝑦=8βˆ’π‘₯.

We now substitute this into the second equation to obtain π‘₯+(8βˆ’π‘₯)=50.

Expanding the brackets and simplifying yields, we get π‘₯+64βˆ’16π‘₯+π‘₯=502π‘₯βˆ’16π‘₯+14=0.

We can note that all three terms share a factor of 2, so we can divide the equation through by 2 to get π‘₯βˆ’8π‘₯+7=0.

We can now solve this equation by factoring. We want two numbers whose product is 7 and whose sum is βˆ’8. We can note that these numbers are βˆ’1 and βˆ’7, so we can factor the quadratic as (π‘₯βˆ’1)(π‘₯βˆ’7)=0.

Thus, the solutions are π‘₯=1 and π‘₯=7. These are the π‘₯-coordinates of the two points of intersection of the graphs.

We can substitute these values into the equation for 𝑦 to determine the 𝑦-coordinates of these two points: 𝑦=8βˆ’1=7,𝑦=8βˆ’7=1.

Hence, we have found that (7,1) and (1,7) are the coordinates of the points of intersection.

We can verify this by substituting the coordinates of either point into the equations of both graphs. For example, let’s verify that (7,1) is a solution by substituting π‘₯=7 and 𝑦=1 into both equations. For the first equation, we have π‘₯+𝑦=87+1=8.

For the second equation, we have π‘₯+𝑦=507+1=5049+1=50.

Since both equations are satisfied, we can conclude that (7,1) is a point of intersection.

Hence, the set of points of intersection between the graphs is {(7,1),(1,7)}.

In our next example, we will solve a linear–quadratic system of equations by graphing both equations.

Example 6: Solving a System of Linear–Quadratic Equations Graphically

  1. Which of the following shows the graph of 𝑦=βˆ’π‘₯+7 and 𝑦=π‘₯+3π‘₯+2?
  2. Find the solutions of the system of equations 𝑦=βˆ’π‘₯+7 and 𝑦=π‘₯+3π‘₯+2.

Answer

Part 1

To sketch the line 𝑦=βˆ’π‘₯+7, we can find the coordinates of two points on the line and connect them with a straight line. We can find two points on the line by substituting.

When π‘₯=0, we have 𝑦=βˆ’0+7=7.

When π‘₯=3, we have 𝑦=βˆ’3+7=4.

Thus, the line 𝑦=βˆ’π‘₯+7 passes through (0,7) and (3,4).

To sketch the quadratic 𝑦=π‘₯+3π‘₯+2, we can start by factoring the right-hand side of the equation. We want to identify two numbers that add to give 3 and multiply to give 2, and we find that these are 1 and 2. Thus, π‘₯+3π‘₯+2=(π‘₯+1)(π‘₯+2).

Therefore, the π‘₯-intercepts of this quadratic are βˆ’1 and βˆ’2. This is a quadratic with a positive leading coefficient, so the shape of the graph will be a parabola that opens upward. We can find the 𝑦-intercept by substituting π‘₯=0 into the equation. We have 𝑦=0+3(0)+2=2.

So, we need to sketch a parabola opening upward with π‘₯-intercepts at βˆ’1 and βˆ’2 and a 𝑦-intercept at 2.

We see that this matches answer C.

Part 2

If a specific pair of values of π‘₯ and 𝑦 satisfies both equations, then we can note that the point with these coordinates must lie on the intersection between both graphs. Therefore, we can solve the equations by finding the coordinates of the points of intersections. We see that these are the points (βˆ’5,12) and (1,6).

Hence, the solutions to the system of equations are π‘₯=βˆ’5, 𝑦=12 and π‘₯=1, 𝑦=6.

In our final example, we solve a real-world problem by sketching a system of equations to find the solutions.

Example 7: Solving a Real-World Problem Involving a System of Equations

A builder wants to construct a triangular girder with a base of length π‘₯ m and a height of length π‘₯βˆ’2 m. The area of this triangle is given by 12π‘₯(π‘₯βˆ’2) m2. If the builder wants the area of the girder to be three times the length of the base, find the length of the base.

Answer

If we call the area of the triangular girder π‘Ž, we note that the area of the triangle is given by the equation π‘Ž=12π‘₯(π‘₯βˆ’2). However, we also want the area to be three times the length of the base, so π‘Ž=3π‘₯.

We can solve this problem by sketching both graphs and finding the coordinates of the points of intersection. First, we know that the graph of π‘Ž=3π‘₯ will be a straight line passing through the origin with slope 3. We can sketch the line by drawing a line through the origin and the point (1,3).

We can sketch the graph of π‘Ž=12π‘₯(π‘₯βˆ’2) by noting that this is a parabola with a positive leading coefficient, so it opens upward. The quadratic is factored, so we can solve each factor equal to zero to find the π‘₯-intercepts. The π‘₯-intercepts are at π‘₯=0 and π‘₯=2, and we can also note that this means the 𝑦-intercept is also at 0.

Sketching these graphs on the same pair of coordinate axes, we find that the coordinates of the points of intersection are (0,0) and (8,24). We cannot have a triangle with a base of length 0, so only the point (8,24) is a valid solution for our triangle.

We can verify this solution by substituting π‘₯=8 into the system of equations. We have π‘Ž=12π‘₯(π‘₯βˆ’2)π‘Ž=12(8)(8βˆ’2)=24,π‘Ž=3π‘₯π‘Ž=3(8)=24.

Since both values for the area are equal, we have verified that π‘₯=8 is the solution to the problem.

Hence, the builder would need to make the base of the triangular girder 8 metres long.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can solve systems of simultaneous equations by finding the coordinates of the points of intersection on the graphs of the equations. In particular, any point on all of the graphs satisfies all of the equations and any solution to the system will be a point of intersection between the graphs.
  • If we can rearrange a linear–quadratic system of equations into the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Žβ‰ 0, then
    • if π‘βˆ’4π‘Žπ‘>0, then the system has 2 solutions;
    • if π‘βˆ’4π‘Žπ‘=0, then the system has 1 solution;
    • if π‘βˆ’4π‘Žπ‘<0, then the system has no real solutions.
  • We can substitute our solutions back into the system to verify that they are in fact solutions to the system.

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