Explainer: Solving Systems of Linear Equations Using Substitution

In this explainer, we will learn how to solve systems of linear equations using substitution.

When we are asked to solve a system of equations (sometimes called simultaneous equations), we are trying to find any points at which the equations are equal, that is, any points where they intersect. Before looking at how we can solve these questions algebraically, let us look at a graphical example.

If we consider the graphs of 𝑦=3π‘₯βˆ’1 and 𝑦=βˆ’2π‘₯+4 and draw them on the same set of axes, we get the following graph.

We can see from this diagram that the graphs intersect at the point (1,2). This, in fact, is the solution to the system of equations 𝑦=3π‘₯βˆ’1 and 𝑦=βˆ’2π‘₯+4. We can check that this is correct by substituting in the value π‘₯=1 into both equations and checking that the corresponding 𝑦-value in each case is 2: 3(1)βˆ’1=2 and βˆ’2(1)+4=2, as required.

Now, to find these solution points algebraically, we have a couple of methods. We can solve the system by trying to eliminate a variable, or we can use substitution. Here, we are going to look at the latter method. This relies on the fact that we are looking for the point where the two equations have equal value which means we can substitute one of our equations into the other. If we look again at the system of equations 𝑦=3π‘₯βˆ’1 and 𝑦=βˆ’2π‘₯+4, to find the value of π‘₯ where the two lines have an equal 𝑦-value, we can substitute one into the other (or equate the two equations); that is 3π‘₯βˆ’1=βˆ’2π‘₯+4. If we then solve this equation, we can find the value of π‘₯ for which the 𝑦-values are equal. First, we add 2π‘₯ to each side of the equation to get 5π‘₯βˆ’1=4.

Then, we add 1 to each side which gives 5π‘₯=5, and then we divide through by 5 to get π‘₯=1.

This means that the two equations have an equal value (that is, the 𝑦-value) when π‘₯=1. To find this value of 𝑦, we can substitute π‘₯ into either one of the equations. Let us substitute into the first: 𝑦=3(1)βˆ’1=2.

Our solutions are, therefore, π‘₯=1 and 𝑦=2 corresponding to the point (1,2), which is the same as we saw from the graph of the two lines.

A few points that are worth highlighting before we start looking at a series of examples are, firstly, that quite often we will need to rearrange one of the equations to get it to a point where we can perform the substitution; secondly, that you do not need to rearrange both equations to the point where you can equate them (although you can); and, thirdly, that you can substitute for either variable. We will address these points more clearly in a few examples.

Example 1: Solving Systems of Linear Equations Using Substitution

Solve the simultaneous equations 𝑦+4π‘₯=βˆ’8 and 𝑦=5π‘₯+10.

Answer

Here, we have a system of two equations. If we call the equation

𝑦+4π‘₯=βˆ’8(1)

and the equation

𝑦=5π‘₯+10,(2)

as (2) is already written in the form 𝑦=…, we can substitute this directly into equation (1). That is, (5π‘₯+10)+4π‘₯=βˆ’8.

If we then simplify the left-hand side by removing the parentheses and collecting like terms, we get 9π‘₯+10=βˆ’8.

Subtracting 10 from each side gives us 9π‘₯=βˆ’18, and dividing through by 9 gives an answer of π‘₯=βˆ’2.

To find the value of 𝑦, we need to substitute into either one of our equations. We will substitute into equation (2) as this is slightly easier: 𝑦=5(βˆ’2)+10=βˆ’10+10=0.

Our solutions are therefore π‘₯=βˆ’2 and 𝑦=0, which corresponds to the point (βˆ’2,0) which is the intersection point of the two lines. A further check that is always worth doing is to substitute your value of π‘₯ into the other equation and check that you get the same 𝑦-value. We will leave that as an exercise for you to do yourself.

Example 2: Using Substitution to Solve Systems of Linear Equations

Solve the simultaneous equations π‘₯βˆ’π‘¦=8 and 3π‘₯βˆ’5𝑦+10=0.

Answer

Our first step in answering this question is to rearrange one of the two equations to make 𝑦 (or π‘₯) the subject. We will call the equation

π‘₯βˆ’π‘¦=8(3)

and the equation

3π‘₯βˆ’5𝑦+10=0.(4)

Here, it is easier to rearrange equation (3). If we add 𝑦 to each side, we get π‘₯=8+𝑦. If we then subtract 8 from each side, we get 𝑦=π‘₯βˆ’8. If we then substitute this equation into the equation (4), we get 3π‘₯βˆ’5(π‘₯βˆ’8)+10=0.

Expanding the parentheses on the left-hand side, we get 3π‘₯βˆ’5π‘₯+40+10=0, and simplifying we get βˆ’2π‘₯+50=0.

Adding 2π‘₯ to each side and dividing through by 2, we find that π‘₯=25.

To find 𝑦, we then need to substitute our value of π‘₯ into our original rearranged equation: 𝑦=(25)βˆ’8=17.

Our solutions are, therefore, π‘₯=25 and 𝑦=17, which corresponds to the point (25,17).

Now, let us look at an example where we eliminate the variable π‘₯ instead of the variable 𝑦. It is often easier to take this approach when solving systems of equations where one of the equations contains an π‘₯-variable with a coefficient of 1 and the 𝑦-variables do not.

Example 3: Solving Systems of Linear Equations Using Substitution

Solve the simultaneous equations π‘₯+4𝑦=17 and 2π‘₯+7𝑦=5.

Answer

If we rearrange the first equation to make π‘₯ the subject, we get π‘₯=17βˆ’4𝑦.

We can then substitute this into the second equation to get 2(17βˆ’4𝑦)+7𝑦=5.

Expanding the parentheses, we get 34βˆ’8𝑦+7𝑦=5, and collecting like terms, we get 34βˆ’π‘¦=5.

If we then add 𝑦 and subtract 5 from each side, we get 𝑦=29.

Finally, we can substitute this into our rearranged equation to get π‘₯: π‘₯=17βˆ’4(29)=17βˆ’116=βˆ’99.

Our solutions are then π‘₯=βˆ’99 and 𝑦=29, which corresponds to the point (βˆ’99,29).

Sometimes, you will be faced with linear equations where you have to do a little more work to rearrange them to make one of the variables the subject. All of the previous examples have contained at least one variable with a coefficient of 1. When this is not the case, you can still use substitution to solve the system which we will demonstrate now, but please be aware that there is an alternate method called elimination which is sometimes easier to use.

Example 4: Solving Systems of Equations Using Substitution

Solve the following system of equations: 4π‘₯+3𝑦=14,5π‘₯+2𝑦=14.

Answer

In this question, we need to choose one of the variables, in either one of the equations, that we want to make the subject. As 2 is the smallest coefficient in either of the two equations, let us make 𝑦 the subject of the second equation. To do this, we first need to subtract 5π‘₯ from both sides to get 2𝑦=14βˆ’5π‘₯.

Then, we need to divide the equation through by 2 which gives us 𝑦=7βˆ’52π‘₯. We can now substitute this into the first equation to get 4π‘₯+3ο€Ό7βˆ’52π‘₯=14.

Expanding the parentheses gives us 4π‘₯+21βˆ’152π‘₯=14.

At this point, we can subtract 21 from both sides to get 4π‘₯βˆ’152π‘₯=βˆ’7.

We now need to collect the like terms on the left-hand side; but before doing this, it is easier to multiply the equation through by 2 to remove the fraction: 8π‘₯βˆ’15π‘₯=βˆ’14. This simplifies to βˆ’7π‘₯=βˆ’14, which means π‘₯=2.

Finally, we can substitute this into our equation for 𝑦: 𝑦=7βˆ’52(2)=7βˆ’5=2.

Our solutions are then π‘₯=2 and 𝑦=2, which corresponds to the point (2,2).

Example 5: Systems of Linear Equations

Solve the following system of equations: 5π‘₯βˆ’2𝑦=8,4π‘₯+3𝑦=11.

Answer

In this question, we need to choose one of the variables, in either one of the equations, that we want to make the subject. As 3 is the smallest positive coefficient in either of the two equations, let us make 𝑦 the subject of the second equation. To do this, we first need to subtract 4π‘₯ from both sides to get 3𝑦=11βˆ’4π‘₯.

Then, we need to divide the equation through by 3 which gives us 𝑦=11βˆ’4π‘₯3.

We can now substitute this into the first equation to get 5π‘₯βˆ’2ο€Ό11βˆ’4π‘₯3=8. Expanding the parentheses gives us 5π‘₯+βˆ’22+8π‘₯3=8.

At this point, it is helpful to multiply the equation through by 3 to remove the fraction, which gives us 15π‘₯βˆ’22+8π‘₯=24. This simplifies to 23π‘₯βˆ’22=24, and adding 22 to each side gives 23π‘₯=46.

If we divide through by 23, we get π‘₯=2. Finally, we can substitute this into our equation for 𝑦:𝑦=11βˆ’4(2)3=11βˆ’83=1.

Our solutions are then π‘₯=2 and 𝑦=1, which corresponds to the point (2,1).

Systems of equations are useful for solving many real-world problems. When presented with a real-world problem, the first thing we need to do is convert the scenario into equations we can solve. In the next example, we will demonstrate this process.

Example 6: Real-World Problems Involving Simultaneous Equations

A movie theater sold 1,200 tickets. If a child’s ticket costs $5.95, an adult’s ticket costs $11.15, and the total revenue was $12,756, how many of each type of ticket did they sell?

Answer

To solve a problem like this, we need to first translate the words into equations. We need to find how many of each type of ticket were sold. Therefore, we need to have a variable for each type of ticket. We will let 𝑐 be the number of child tickets that were sold and π‘Ž be the number of adult tickets that were sold. We know that a total of 1,200 tickets were sold. Therefore, our first equation is

π‘Ž+𝑐=1,200.(5)

We also know the cost of each ticket and the total revenue. Using these facts, we can form a second equation. The total revenue from the sale of child tickets will be given by the cost of a ticket multiplied by the number sold, which we can write mathematically as 5.95𝑐 Similarly, the total revenue from the sale of adult tickets will be 11.15π‘Ž. Hence, the total revenue for both, which we know is $12,756, will be the sum of these. Hence, our second equation is

5.95𝑐+11.15π‘Ž=12,756.(6)

Using equation (5), we can get an expression for π‘Ž by subtracting 𝑐 from both sides of the equation: π‘Ž=1,200βˆ’π‘.

We can now substitute this into equation (6) to get 5.95𝑐+11.15(1,200βˆ’π‘)=12,756.

Expanding the parentheses, we have 5.95𝑐+11.15Γ—1,200βˆ’11.15𝑐=12,756.

Gathering together our terms in 𝑐 and simplifying, we have 13,380βˆ’5.2𝑐=12,756.

Adding 5.2⋅𝑐 to both sides of the equation yields 13,380=12,756+5.2𝑐.

We can now subtract 12,756 from both sides to get 5.2𝑐=624.

Finally, we can divide both sides of the equation by 5.2 to get 𝑐=120.

We can now substitute this value of 𝑐 into π‘Ž=1,200βˆ’π‘ to find the value of π‘Ž as follows: π‘Ž=1,200βˆ’120=1,080.

At this stage, it is good practice to substitute these values into the other equation to ensure we have correctly done our working; substituting π‘Ž=1,080 and 𝑐=120 into the left-hand side of equation (6) gives 5.95(120)+11.15(1,080)=714+12,042=12,756

as expected. Hence, the number of child tickets sold was 120 and the number of adult tickets sold was 1,080.

Key Points

  • Systems of linear equations can be solved using different techniques.
  • To solve a system of linear equations using substitution, we use the following method:
    1. Rearrange one of the equations to make one of the unknowns the subject.
    2. Substitute this into the other equation and solve for the unknown.
    3. Substitute the value of this unknown into one of the equations and solve for the remaining unknown.
    4. Check that the values of both unknowns satisfy the other equation.

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