Lesson Explainer: Solving Systems of Linear Equations by Omitting a Variable | Nagwa Lesson Explainer: Solving Systems of Linear Equations by Omitting a Variable | Nagwa

Lesson Explainer: Solving Systems of Linear Equations by Omitting a Variable Mathematics • 8th Grade

In this explainer, we will learn how to solve systems of linear equations by omitting a variable.

When we are asked to solve a system of equations, this means we are looking for a set of values for the variables that satisfy every equation. For example, consider the system of equations π‘₯+𝑦=1,π‘₯βˆ’π‘¦=3.

We want to find a value for π‘₯ and a value for 𝑦 such that both equations hold true. In other words, we are looking for two values whose sum is 1 and whose difference is 3. We could do this by trial and error; however, this will not work for more complicated systems.

Instead, we will use the fact that we can solve any linear equation in one variable. This means if we can find a linear equation in either variable, we can solve for that value. To do this, we can note that both equations must hold true, so both sides of the equation are equal (for both equations). This means we can add the equations together: π‘₯+𝑦=1+(π‘₯βˆ’π‘¦=3)2π‘₯+0𝑦=4.

We add the left-hand side of both equations to get 2π‘₯, and we add the right-hand side of both equations to get 4. Hence, 2π‘₯=4, which we can solve by dividing through by 2, giving us π‘₯=2.

Finally, we substitute π‘₯=2 into the first equation to get 2+𝑦=1, and we then subtract 2 from both sides of the equation to get 𝑦=βˆ’1.

Therefore, π‘₯=2 and 𝑦=βˆ’1 is a solution to this system of equations. We can check that this solution is correct by substituting these two values back into the system of equations to check that each equation is valid; though it is more efficient to only substitute into the equation we did not use to find the values. We have π‘₯+𝑦=2+(βˆ’1)=1,π‘₯βˆ’π‘¦=2βˆ’(βˆ’1)=3.

This agrees with the system of equations, so this confirms that the solution is correct.

To find this solution, we added the equations together; however, this only worked because the terms 𝑦 and βˆ’π‘¦ canceled out to leave us with a linear equation in π‘₯. This is why the process is called omission, since we omit one of the variables. This is not the only way we could have omitted a variable from these equations; we could also subtract the second equation from the first, giving us π‘₯+𝑦=1βˆ’(π‘₯βˆ’π‘¦=3)0π‘₯+2𝑦=βˆ’2.

We can then solve for 𝑦 to get 𝑦=βˆ’1 and then substitute 𝑦=βˆ’1 into the first equation to see that π‘₯+(βˆ’1)=1, so π‘₯=2.

In both of these cases, we were able to omit a variable by using the fact that the magnitudes of the coefficients of one of the variables in both equations were equal. In other words, since the coefficient of π‘₯ was 1 in both equations, we could omit π‘₯ by subtracting the equations. Similarly, since the coefficients of 𝑦 were 1 and βˆ’1 in each equation, which have the same magnitude but opposite sign, we could add the equations to omit 𝑦.

This will not always be the case, so let’s see how to apply this method to solve the following system of linear equations: 2π‘₯+𝑦=1,π‘₯βˆ’3𝑦=1.

We see that the magnitudes of the coefficients of both unknowns are not equal, so we cannot just add or subtract the equations to omit a variable. Instead, we will need to rewrite one of the equations so that this is true. We will multiply the second equation through by 2 to get 2π‘₯βˆ’6𝑦=2. Now, we have the following system of linear equations: 2π‘₯+𝑦=1,2π‘₯βˆ’6𝑦=2.

We can subtract the second equation from the first to omit the variable π‘₯; this gives 2π‘₯+𝑦=1βˆ’(2π‘₯βˆ’6𝑦=2)7𝑦=βˆ’1.

Dividing through by 7 gives 𝑦=βˆ’17.

We can substitute this value into the first equation to see that 2π‘₯βˆ’17=1.

We can then solve this equation for π‘₯ by adding 17 to both sides of the equation: 2π‘₯=87.

Finally, we divide through by 2 to get π‘₯=47.

It is worth noting that we also could have multiplied the first equation through by 3 and then added the equations to omit the variable 𝑦.

Before we move onto examples, there is one further type of system of two linear equations that can occur. In these systems, the coefficients of the variables are not direct multiples of each other, so we would need to multiply both equations to omit a variable. Let’s see an example of this in the following system of equations: 2π‘₯+3𝑦=10,3π‘₯βˆ’4𝑦=βˆ’2.

We note that the coefficients of the same variable are not direct multiples of each other, so instead, we will multiply the first equation by 3 and the second equation by 2 so that the π‘₯-coefficients have the same magnitude. This gives 3Γ—(2π‘₯+3𝑦)=3Γ—106π‘₯+9𝑦=30,2Γ—(3π‘₯βˆ’4𝑦)=2Γ—(βˆ’2)6π‘₯βˆ’8𝑦=βˆ’4.

So, we have rewritten the system as 6π‘₯+9𝑦=30,6π‘₯βˆ’8𝑦=βˆ’4.

We can then subtract the second equation from the first to omit the π‘₯-variable: 6π‘₯+9𝑦=30βˆ’(6π‘₯βˆ’8𝑦=βˆ’4)17𝑦=34.

We can then find the value of 𝑦 by dividing through by 17 to get 𝑦=3417=2.

Finally, we substitute 𝑦=2 into the first equation to find π‘₯: 2π‘₯+3(2)=102π‘₯=4π‘₯=2.

Let’s now summarize how to use the omission method to solve systems of two linear equations.

How To: Solving a System of Two Linear Equations in Two Variables by Omission

The steps for solving a system of two linear equations using omission are as follows:

  1. Identify whether the system contains a pair of coefficients of either unknown that have equal magnitude.
  2. If the system does not have a pair of coefficients of equal magnitude, multiply one or both equations through by a constant to make a pair of coefficients have equal magnitude.
  3. Once the system has a pair of coefficients with equal magnitude, subtract the equations if the coefficients are the same and add the equations if the coefficients are of opposite signs to omit the variable.
  4. Solve the resulting linear equation to find one of the unknowns.
  5. Substitute this value back into one of the original equations to find the other unknown.

Let’s now see some examples of how to apply the omission method to solve systems of equations.

Example 1: Solving Simultaneous Linear Equations with 𝑦-Coefficients Having Equal Magnitude but Opposite Signs

Using omission, solve the simultaneous equations 3π‘₯+2𝑦=14,6π‘₯βˆ’2𝑦=22.

Answer

To solve simultaneous equations using omission, we want to add or subtract a multiple of the equations to omit a variable. In this case, we see that we can add the 𝑦-terms in each equation to omit the variable 𝑦, so we add the equations together: 3π‘₯+2𝑦=14+6π‘₯βˆ’2𝑦=229π‘₯=36.

We can then divide the resulting equation through by 9 to find π‘₯. We get π‘₯=369=4.

Now, we substitute π‘₯=4 into the original equation to find the value of 𝑦 to get 3(4)+2𝑦=1412+2𝑦=14.

Next, we subtract 12 from both sides of the equation to find that 2𝑦=2.

Finally, we divide through by 2 to see that 𝑦=1.

Hence, the solution to the system of equations is π‘₯=4, 𝑦=1.

Example 2: Solving Simultaneous Equations by Omission

Use the omission method to solve the simultaneous equations 3π‘Ž+2𝑏=14,4π‘Ž+2𝑏=16.

Answer

To solve simultaneous equations by using omission, we want to add or subtract a multiple of the equations to omit a variable. In this case, we see that the difference in the 𝑏-terms in each equation is 0. Hence, we can subtract the two equations to omit the 𝑏-variable, which gives 3π‘Ž+2𝑏=14βˆ’(4π‘Ž+2𝑏=16)βˆ’π‘Ž=βˆ’2.

We can then multiply the resulting equation through by βˆ’1 to find π‘Ž, and we get π‘Ž=2.

Now, we substitute π‘Ž=2 into the original equation to find the value of 𝑏. We get 3π‘Ž+2𝑏=146+2𝑏=14.

Next, we subtract 6 from both sides of the equation to find that 2𝑏=8.

Finally, we divide through by 2 to see that 𝑏=4.

This is enough to answer our question; however, we can also check our answer by substituting these values back into the system of equations to check that both equations are satisfied.

Substituting π‘Ž=2 and 𝑏=4 into the left-hand side of the first equation yields 3(2)+2(4)=6+8=14, which is equal to the right-hand side of the first equation.

Substituting π‘Ž=2 and 𝑏=4 into the left-hand side of the second equation yields 4(2)+2(4)=8+8=16, which is equal to the right-hand side of the second equation. Since both equations hold true, this verifies that our solution is correct.

Hence, the solution to the simultaneous equations is π‘Ž=2, 𝑏=4.

In our next example, we will need to multiply one of the equations by a constant in order to omit a variable.

Example 3: Solving Simultaneous Equations Using Omission, Where One of the Equations Needs to be Multiplied

Using omission, solve the simultaneous equations 5π‘₯+4𝑦=27,3π‘₯+12𝑦=45.

Answer

To omit one of the variables in this equation, we need the magnitudes of the coefficients of a single variable to be equal in both equations. We can do this by noting that 3Γ—4𝑦=12𝑦, so we can multiply the first equation through by 3 to omit the variable 𝑦. This gives 3Γ—(5π‘₯+4𝑦)=3Γ—2715π‘₯+12𝑦=81.

So, the simultaneous equations are now 15π‘₯+12𝑦=81,3π‘₯+12𝑦=45.

We can then subtract one equation from the other to omit 𝑦: 15π‘₯+12𝑦=81βˆ’(3π‘₯+12𝑦=45)12π‘₯=36.

We then divide through by 12 to find π‘₯, and we see that π‘₯=3.

Now, we substitute π‘₯=3 into the first equation and solve to find 𝑦 to get 5(3)+4𝑦=2715+4𝑦=27.

Then, we subtract 15 from both sides of the equation to get 4𝑦=12.

Finally, we divide both sides of the equation through by 4, giving us 𝑦=3.

Hence, the solution to the simultaneous equations is π‘₯=3, 𝑦=3.

In our next example, in order to omit a variable from the simultaneous equations, we will need to multiply both equations by a constant.

Example 4: Solving Simultaneous Equations Using Omission, Where Both of the Equations Need to be Multiplied

Using omission, solve the simultaneous equations 4π‘₯+6𝑦=40,3π‘₯+7𝑦=40.

Answer

To omit one of the variables in this equation, we need the magnitudes of the coefficients of a single variable to be equal in both equations. We cannot do this by multiplying a single equation by a constant since the coefficients of the same variable are not integer multiples of each other.

So instead, we will multiply both equations so that the π‘₯-coefficients are equal. We will multiply the first equation through by 3 and the second equation through by 4. This gives us 12π‘₯+18𝑦=120,12π‘₯+28𝑦=160.

We can now subtract one equation from the other to omit the variable π‘₯; this gives 12π‘₯+18𝑦=120βˆ’(12π‘₯+28𝑦=160)βˆ’10𝑦=βˆ’40.

Now, we divide the resulting equation through by βˆ’10 to solve for 𝑦. We get 𝑦=βˆ’40βˆ’10=4.

We can then substitute 𝑦=4 into the first equation to construct a linear equation for π‘₯ to get 4π‘₯+6(4)=404π‘₯+24=40.

We subtract 24 from both sides, giving us 4π‘₯=16.

Finally, we divide both sides of the equation through by 4 to get π‘₯=4.

This is enough to answer our question; however, we can also check our answer by substituting these values back into the system of equations to check that both equations are satisfied.

Substituting π‘₯=4 and 𝑦=4 into the left-hand side of the first equation yields 4(4)+6(4)=16+24=40, which is equal to the right-hand side of the first equation.

Substituting π‘₯=4 and 𝑦=4 into the left-hand side of the second equation yields 3(4)+7(4)=12+28=40, which is equal to the right-hand side of the second equation. Since both equations hold true, this verifies that our solution is correct.

Hence, the solution to the simultaneous equations is π‘₯=4, 𝑦=4.

In our final example, we will apply the omission method to solve a geometric problem involving a system of linear equations.

Example 5: Solving a System of Linear Equations to Find Unknown Side Lengths in a Triangle Using the Perimeter

𝐴𝐡𝐢 is a triangle, where 𝐡𝐢=55cm, π΄πΆβˆ’π΄π΅=13cm, and the perimeter is 124 cm. Find the lengths of 𝐴𝐢 and 𝐴𝐡 giving the answers to the nearest centimetre.

Answer

We first recall that the perimeter of a polygon is the sum of all of its side lengths. Since 𝐴𝐡𝐢 is a triangle, its perimeter is the sum of the lengths of its three sides. We are told the perimeter of this triangle is 124 cm, so we have 𝐴𝐡+𝐡𝐢+𝐴𝐢=124.

We are also told that 𝐡𝐢=55. Substituting this into the equation for the perimeter, we get 𝐴𝐡+55+𝐴𝐢=124.

We can subtract 55 from both sides to get 𝐴𝐡+𝐴𝐢=69.

We are also given that π΄πΆβˆ’π΄π΅=13.

Therefore, we have two equations in two unknowns; we can attempt to solve this by omitting one of the unknowns. By reordering the second equation, we have the following pair of simultaneous equations: 𝐴𝐡+𝐴𝐢=69,βˆ’π΄π΅+𝐴𝐢=13.

We note that adding the equations will omit 𝐴𝐡 from the equation, so we add the equations together to get 𝐴𝐡+𝐴𝐢=69+βˆ’π΄π΅+𝐴𝐢=132𝐴𝐢=82.

Now, we divide the resulting equation through by 2 to get 𝐴𝐢=41.cm

Substituting 𝐴𝐢=41 into the equation 𝐴𝐡+𝐴𝐢=69 yields 𝐴𝐡+41=69.

We subtract 41 from both sides of the equation to find 𝐴𝐡=28.cm

Hence, if 𝐴𝐡𝐢 is a triangle, where 𝐡𝐢=55cm, π΄πΆβˆ’π΄π΅=13cm, and the perimeter is 124 cm, then 𝐴𝐢=41cm and 𝐴𝐡=28cm.

Before we finish the explainer, it is worth pointing out that it is not always possible to solve systems of two linear equations using omission. For example, imagine we are told that the sum of two numbers is three and that the sum of two numbers is two. Of course, we know this is not possible, but we can still construct a system of equations representing each sum. We get π‘₯+𝑦=3,π‘₯+𝑦=2.

If we were to try and solve this system by omission, we would subtract the equations to get π‘₯+𝑦=3βˆ’(π‘₯+𝑦=2)0=1.

We know that zero is not equal to one; in fact, we can also say that no values of π‘₯ and 𝑦 will make zero equal to one. In other words, there are no values of π‘₯ and 𝑦 that satisfy this equation.

It is also worth noting that there can be an infinite number of solutions. For example, consider the following system: 2π‘₯+2𝑦=2,π‘₯+𝑦=1.

We can attempt to solve this by using omission; we multiply the second equation through by 2 to get 2π‘₯+2𝑦=2.

This is then exactly the same as the first equation, so when we omit the variable, we get 2π‘₯+2𝑦=2βˆ’(2π‘₯+2𝑦=2)0=0.

We then note that zero is always equal to zero, for any values of π‘₯ and 𝑦. Hence, any value of π‘₯ will give a corresponding value of 𝑦 that solves the equation; there are an infinite number of solutions to the system. We will not go into full detail on these cases since they are outside of the scope of the explainer, but it is important to note that these cases exist.

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

Key Points

  • The steps for solving a system of two linear equations using omission are as follows:
    1. Identify whether the system contains a pair of coefficients of either unknown that have equal magnitude.
    2. If the system does not have a pair of coefficients of equal magnitude, multiply one or both equations through by a constant to make a pair of coefficients have equal magnitude.
    3. Once the system has a pair of coefficients with equal magnitude, add or subtract the equations to omit the variable.
    4. Solve the resulting equation to find one of the unknowns.
    5. Substitute this value back into one of the original equations to find the other unknown.
  • Solving a pair of simultaneous equations by using omission gives us exact solutions.
  • We can verify our solutions by substituting them back into the system of equations to check that the equations hold.

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