Lesson Explainer: Linear–Quadratic Systems of Equations Mathematics

In this explainer, we will learn how to solve linear–quadratic systems of equations.

Systems of equations appear throughout many areas of science including finance, computing, and mechanics. To solve a system of equations means to find values for the variables that make every equation true.

How To: Finding the Solutions to a System of Equations by Substitution

For example, if we are given the equations π‘₯+𝑦=2,2π‘₯+𝑦=3, then, by inspection, we can see that π‘₯=1 and 𝑦=1 is a solution. By substituting these values into each equation, we get π‘₯+𝑦=21+1=22=2,2π‘₯+𝑦=32(1)+1=33=3.

Since both equations hold true, we have confirmed that a solution to this system is π‘₯=1 and 𝑦=1. We can find this solution from the equations in a number of different ways, such as elimination, substitution, and graphing to name a few. In this explainer, we will focus on the substitution method.

To solve this system of equations by substitution, we need to rearrange one of the equations to find an expression for one of our variables. One way of doing this is to rearrange the first equation as follows: π‘₯+𝑦=2π‘₯+π‘¦βˆ’π‘¦=2βˆ’π‘¦π‘₯=2βˆ’π‘¦.

Hence, for a pair of values π‘₯ and 𝑦 to solve this system, we must have π‘₯=2βˆ’π‘¦. We can then substitute this expression for π‘₯ into the other equation that must hold true as follows: 2π‘₯+𝑦=32(2βˆ’π‘¦)+𝑦=3.

We can then solve this equation for 𝑦 as follows: 2(2βˆ’π‘¦)+𝑦=34βˆ’2𝑦+𝑦=34βˆ’π‘¦=34=3+𝑦1=𝑦.

Therefore, 𝑦=1 in any solution to this system of equations. We also need to find the value π‘₯; we can do this by substituting this value for 𝑦 into either equation. This gives us π‘₯+𝑦=2π‘₯+1=2π‘₯=2βˆ’1π‘₯=1.

Hence, the solution to this system of equations is π‘₯=1 and 𝑦=1. We could then confirm our solution by substituting these values back into the system of equations. We can see why this is true by sketching the graphs of both equations.

The point of intersection of the graphs will be the point where both equations are satisfied. We can see that there is only one point of intersection at (1,1), confirming our solution is correct.

The system of equations in the example above is called a first-degree system of equations. This is because every equation is linear (in fact, every equation is a first-degree polynomial). In the remainder of the explainer, we will focus on finding solutions to second-degree systems, which we will define below.

Definition: Linear–Quadratic System of Equations

A linear–quadratic system of equations is a system of equations formed of exactly one first-degree polynomial equation and one second-degree polynomial equation.

To fully understand this, we need to understand what is meant by a polynomial in two variables and how to determine its degree.

Definition: Polynomial Functions in Two variables

A polynomial in two variables is a function in which every term is a monomial. In particular, the variables must have nonnegative integer exponents.

The degree of a polynomial is the greatest sum of the degrees of the variables in a single term.

For example, this means that equations like 𝑦+π‘₯=1,π‘₯𝑦=12,π‘₯=𝑦+3𝑦+1,𝑦=2π‘₯+1, can appear in second-degree systems of equations since they are all polynomial of degree at most 2. We can find the degree of each of these polynomials as follows:

  • 𝑦+π‘₯=1 is a second-degree polynomial since the terms π‘₯ and π‘¦οŠ¨ are power two terms.
  • π‘₯𝑦=12 is also a second-degree polynomial; we need to take the sum of the powers of the variables. We see that π‘₯𝑦=π‘₯π‘¦οŠ§οŠ§ and 1+1=2, so this polynomial has degree 2.
  • π‘₯=𝑦+3𝑦+1 is a second-degree polynomial since it is a quadratic in the variable 𝑦.
  • 𝑦=2π‘₯+1 is a first-degree polynomial since it is linear.

It is also worth noting that we refer to degree 2 polynomials in two variables as quadratics in two variables. For example, π‘Žπ‘₯+𝑏π‘₯𝑦+𝑐𝑦+𝑑π‘₯+𝑒𝑦+𝑓=0 is a quadratic equation in two variables, for real constants π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, and 𝑓.

We will now discuss how to solve a second-degree system of equations where one of the given equations is linear.

How To: Finding the Solutions to a Linear–Quadratic System of Equations by Substitution

Let’s try to solve the linear–quadratic system of equations π‘₯+𝑦=4,π‘₯+𝑦=10.

Since one of the given equations is linear, we can find an expression for 𝑦 in terms of π‘₯ by rearranging as follows: π‘₯+𝑦=4π‘₯+π‘¦βˆ’π‘₯=4βˆ’π‘₯𝑦=4βˆ’π‘₯.

We can then substitute this expression for 𝑦 into the nonlinear equation and simplify as follows: π‘₯+𝑦=10π‘₯+(4βˆ’π‘₯)=10π‘₯+16βˆ’4π‘₯βˆ’4π‘₯+π‘₯=102π‘₯βˆ’8π‘₯+16=102π‘₯βˆ’8π‘₯+6=0.

We can take out the shared factor of 2 to get 2ο€Ήπ‘₯βˆ’4π‘₯+3=0π‘₯βˆ’4π‘₯+3=0.

This is then a quadratic equation in π‘₯; one way of solving this equation is by factoring. We need to find two numbers that multiply to give 3 and add to give βˆ’4. We note that (βˆ’3)Γ—(βˆ’1)=3 and (βˆ’3)+(βˆ’1)=βˆ’4, so we can factor the quadratic as (π‘₯βˆ’3)(π‘₯βˆ’1)=0.

For a product to be equal to zero, one of the factors must be equal to zero; hence, we have π‘₯βˆ’3=0π‘₯βˆ’1=0.or

Solving the first equation, we have π‘₯=3; solving the second equation, we have π‘₯=1. We can find the corresponding 𝑦-values by substituting these values into the linear equation.

First, we substitute π‘₯=3 into the equation, giving us π‘₯+𝑦=43+𝑦=4𝑦=1.

Then, we substitute π‘₯=1 into the equation, giving us π‘₯+𝑦=41+𝑦=4𝑦=3.

Hence, we have found two different solutions, either π‘₯=3 and 𝑦=1 or π‘₯=1 and 𝑦=3. It is worth noting these solutions only work in pairs. π‘₯=3 and 𝑦=1 is one solution to the system of equations. Similarly, π‘₯=1 and 𝑦=3 is another solution; we cannot have π‘₯=1 and 𝑦=1 together as this will not solve the system.

We can verify both solutions by substituting them into the system of equations and checking that the equations are satisfied.

In our first example, we will determine all of the solutions of a pair of simultaneous equations where one equation is linear and the other is nonlinear.

Example 1: Solving Systems of Linear and Quadratic Equations

Given that π‘₯+π‘₯𝑦=18 and π‘₯+𝑦=6, find the value of π‘₯.

Answer

We need to find the value of π‘₯ that satisfies this equation. We can do this by finding the pairs of values of π‘₯ and 𝑦 that satisfy both equations. We will do this by using substitution. We could start by rearranging the linear equation to find an expression for either of the variables π‘₯ and 𝑦. However, before we do this, we can look at the nonlinear equations we are given: π‘₯+π‘₯𝑦=18. In this equation, the highest power of 𝑦 is one, but the highest power of π‘₯ is 2; this means it will be easier to substitute an expression for 𝑦, since we will not have to square a binomial.

So, we start by finding an expression for 𝑦: π‘₯+𝑦=6𝑦=6βˆ’π‘₯.

We then substitute this expression into the nonlinear equation and simplify as follows: π‘₯+π‘₯𝑦=18π‘₯+π‘₯(6βˆ’π‘₯)=18π‘₯+6π‘₯βˆ’π‘₯=186π‘₯=18.

We can then solve for π‘₯ by dividing through by 6. This gives us π‘₯=186=3.

Hence, if π‘₯+π‘₯𝑦=18 and π‘₯+𝑦=6, then π‘₯=3.

In our next two examples, we will solve a system of equations to determine the coordinates of the points of intersection of two curves.

Example 2: Solving Systems of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs

Find the set of points of intersection of the graphs of π‘₯+7=8 and π‘₯+𝑦=65.

Answer

We recall that the graphs of two equations will intersect when their π‘₯- and 𝑦-coordinates are equal; hence, we need to find the values of π‘₯ and 𝑦 that satisfy both equations.

There are many different methods for solving simultaneous equations; since we are given one linear equation and one nonlinear equation, we will use substitution. To use substitution, we need to find an expression for one of the variables from the linear equation. We have π‘₯+7=8π‘₯=8βˆ’7π‘₯=1.

We can then substitute this value for π‘₯ into the nonlinear equation and simplify to get π‘₯+𝑦=651+𝑦=65𝑦=64.

We then take the square root of both sides of the equation, where we note we will get a positive and negative solution as follows: 𝑦=Β±8.

Hence, the set of points of intersection of the graphs of π‘₯+7=8 and π‘₯+𝑦=65 is {(1,8),(1,βˆ’8)}.

Example 3: Solving Systems of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Graphs

Find the set of points of intersection of the graphs of π‘₯+5𝑦=0 and 𝑦=βˆ’π‘₯.

Answer

For a point to be on both graphs, its coordinates must satisfy both equations. Therefore, to determine the set of points of intersection of the given equations, we need to solve the simultaneous equations π‘₯+5𝑦=0,𝑦=βˆ’π‘₯.

Since one of the given equations is linear, we will attempt to solve these equations by substitution. We start by rearranging the linear equation to make π‘₯ the subject as follows: π‘₯+5𝑦=0π‘₯=βˆ’5𝑦.

We then substitute this expression for π‘₯ into the other equation and simplify as follows: 𝑦=βˆ’π‘₯𝑦=βˆ’(βˆ’5𝑦)𝑦=5𝑦.

We can then solve this equation by rearranging and factoring as follows: π‘¦βˆ’5𝑦=0𝑦(π‘¦βˆ’5)=0.

For a product to be equal to zero, one of the factors must be equal to zero; hence, we have 𝑦=0𝑦=5.or

To find the π‘₯-coordinates of the points of intersection, we can substitute the 𝑦-coordinate into the equation for either curve. Since π‘₯+5𝑦=0 is a linear equation, it will be easier to use this equation.

First, we substitute 𝑦=0 into the equation, giving us π‘₯+5(0)=0π‘₯=0.

Then, we substitute 𝑦=5 into the equation, giving us π‘₯+5(5)=0π‘₯+25=0π‘₯=βˆ’25.

This gives us two points of intersection: (0,0) and (βˆ’25,5).

Finally, we can check that these solutions are correct by substituting them into the simultaneous equations. For example, we have π‘₯=βˆ’25 and 𝑦=5, which we substitute into the following equations respectively: 𝑦=βˆ’π‘₯5=βˆ’(βˆ’25)25=25,π‘₯+5𝑦=0βˆ’25+5(5)=0.

Since both equations hold true for these values of π‘₯ and 𝑦, we have confirmed this is a solution of the simultaneous equations. We could also check the solution π‘₯=0 and 𝑦=0 in the same way.

Since the questions asks for the set of points of intersection, we write these in a set as {(0,0),(βˆ’25,5)}.

In our next two examples, we will solve a pair of simultaneous equations that have multiple solutions, where one equation is linear and the other is nonlinear.

Example 4: Solving Systems of Nonlinear Equations

Solve the simultaneous equations 𝑦=π‘₯βˆ’4,π‘₯5+𝑦3=4, giving your answers to two decimal places.

Answer

We can attempt to solve a system of equations by using substitution. The linear equation is already given with 𝑦 as the subject, so we will substitute this expression into the nonlinear equation, giving us π‘₯5+𝑦3=4π‘₯5+(π‘₯βˆ’4)3=4.

To solve this equation for π‘₯, we need to distribute and simplify. We can start by distributing the square (π‘₯βˆ’4)3=π‘₯βˆ’8π‘₯+163.

We can then use this to simplify the equation as follows: π‘₯5+π‘₯βˆ’8π‘₯+163=4π‘₯5Γ—33+π‘₯βˆ’8π‘₯+163Γ—55=43π‘₯15+5π‘₯βˆ’40π‘₯+8015=43π‘₯+5π‘₯βˆ’40π‘₯+8015=48π‘₯βˆ’40π‘₯+80=608π‘₯βˆ’40π‘₯+20=0.

We can then divide through by 4 to get 2π‘₯βˆ’10π‘₯+5=0.

We can solve this by using the quadratic formula that states that the solutions to the quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0 are π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, provided the discriminant is nonnegative and π‘Žβ‰ 0.

We have π‘Ž=2, 𝑏=βˆ’10, and 𝑐=5. Substituting these values in gives us π‘₯=βˆ’(βˆ’10)±(βˆ’10)βˆ’4(2)(5)2(2)=10±√100βˆ’404=10Β±2√154=5±√152.

Evaluating these two roots to two decimal places, we have π‘₯β‰ˆ0.56 and π‘₯β‰ˆ4.44.

We can find the values of 𝑦 by recalling that 𝑦=π‘₯βˆ’4. Hence, 𝑦=5±√152βˆ’4=βˆ’3±√152.

Calculating 𝑦 to two decimal places, we have π‘₯β‰ˆ0.56 and π‘¦β‰ˆβˆ’3.44 and π‘₯β‰ˆ4.44 and π‘¦β‰ˆ0.44.

Example 5: Solving a System of Linear and Quadratic Equations

Find all of the solutions to the simultaneous equations 𝑦+9π‘₯+7=0 and π‘₯+π‘¦βˆ’7π‘₯𝑦=4. Give values to two decimal places.

Answer

There are many different methods for solving simultaneous equations; since we are given one linear equation and one nonlinear equation, we will use substitution. To use substitution, we need to find an expression for one of the variables from the linear equation. We have 𝑦+9π‘₯+7=0𝑦=βˆ’9π‘₯βˆ’7.

We can then substitute this expression for 𝑦 into the nonlinear equation as follows: π‘₯+π‘¦βˆ’7π‘₯𝑦=4π‘₯+(βˆ’9π‘₯βˆ’7)βˆ’7π‘₯(βˆ’9π‘₯βˆ’7)=4.

We need to then solve this equation for π‘₯; we can do this by first distributing over the parentheses. We will distribute each separately.

First, we have (βˆ’9π‘₯βˆ’7)=81π‘₯+126π‘₯+49.

Second, we have βˆ’7π‘₯(βˆ’9π‘₯βˆ’7)=63π‘₯+49π‘₯.

Substituting these expressions into the equation and simplifying gives us π‘₯+(βˆ’9π‘₯βˆ’7)βˆ’7π‘₯(βˆ’9π‘₯βˆ’7)=4π‘₯+81π‘₯+126π‘₯+49+63π‘₯+49π‘₯=4145π‘₯+175π‘₯+45=0.

We can simplify by dividing through by 5. This gives us 29π‘₯+35π‘₯+9=0.

This is then a quadratic equation in π‘₯. We can solve this by applying the quadratic formula that states that the solutions to the quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0 are π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, provided the discriminant is nonnegative and π‘Žβ‰ 0. In this quadratic equation π‘Ž=29, 𝑏=35, and 𝑐=9. Substituting these values in, we have π‘₯=βˆ’35±√35βˆ’4(29)(9)2(29)=βˆ’35±√18158.

Evaluating both roots to two decimal places, we have π‘₯β‰ˆβˆ’0.84 and π‘₯β‰ˆβˆ’0.37. However, we will need to use the exact values to determine the values of 𝑦. We do this by substituting the π‘₯-values into the equation 𝑦=βˆ’9π‘₯βˆ’7.

This gives us 𝑦=βˆ’9ο€Ώβˆ’35+√18158ο‹βˆ’7𝑦=βˆ’9ο€Ώβˆ’35βˆ’βˆš18158ο‹βˆ’7.

We can calculate these to two decimal places to get π‘₯β‰ˆβˆ’0.37 and π‘¦β‰ˆβˆ’3.66 and π‘₯β‰ˆβˆ’0.84 and π‘¦β‰ˆ0.52. We can write these solution pairs a set, giving us {(βˆ’0.37,βˆ’3.66),(βˆ’0.84,0.52)}.

In our final example, we will see how a real-world problem can involve solving a linear–quadratic system of equations.

Example 6: Solving Systems of Linear and Quadratic Equations

A father’s age is 10 more than 2 times his son’s age. The sum of the squares of their ages is 4 more than 3 times the product of their ages. What are their ages?

Answer

Let’s start by forming equations from the given information. Let’s call the father’s age 𝑦 and the son’s age π‘₯. Since the father’s age is 10 more than 2 times his son’s age, we must have 𝑦=2π‘₯+10.

Next, since the sum of the squares of their ages is 4 more than 3 times the product of their ages, we must also have that π‘₯+𝑦=3π‘₯𝑦+4.

This is a pair of simultaneous equations where one is linear equation and the other is nonlinear. We can solve this by using substitution. We start by substituting the expression for 𝑦 into the nonlinear equation, which gives us π‘₯+(2π‘₯+10)=3π‘₯(2π‘₯+10)+4.

This is now an equation in π‘₯. We can solve this by first distributing the parentheses and simplifying as follows: π‘₯+(2π‘₯+10)=3π‘₯(2π‘₯+10)+4π‘₯+4π‘₯+40π‘₯+100=6π‘₯+30π‘₯+4π‘₯βˆ’10π‘₯βˆ’96=0.

We can solve this by factoring. We note that (βˆ’16)Γ—6=βˆ’96 and βˆ’16+6=βˆ’10, so we can factor the quadratic equation to get (π‘₯βˆ’16)(π‘₯+6)=0.

Solving each factor to be equal to zero, we get two solutions to the simultaneous equations: π‘₯=βˆ’6 and π‘₯=16. However, π‘₯ is the son’s age in years, so π‘₯ cannot be negative. Hence, there is only one solution when π‘₯=16.

We can find the father’s age by substituting π‘₯=16 into the equation 𝑦=2π‘₯+10, which gives us 𝑦=2(16)+10=42.

Hence, the father is 42 years old and the son is 16 years old. We can note that these ages seem reasonable for the ages of a father and son, and we can check further by substituting π‘₯=16 and 𝑦=42 back into the simultaneous equations. This gives us 𝑦=2π‘₯+1042=2(16)+1042=42,π‘₯+𝑦=3π‘₯𝑦+416+42=3(16)(42)+42020=2020.

Since both equations are satisfied, we can conclude these are the solutions. Hence, the father’s age is 42 years, the son’s age is 16 years.

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

Key Points

  • A linear–quadratic system of equations is a system of equations formed of a linear equation and a quadratic equation.
  • A linear–quadratic system of equations can have two solutions, one solution, or zero solutions.
  • We can find the solutions to these systems of equations by rearranging the linear equation to find an expression for a variable and then substituting the expression into the nonlinear equation.
  • We can use this method to find the coordinates of the points of intersection of some graphs.

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