Linear Equations in Three Variables

To solve a systems of linear equations in three variables x, y, and z, we need to use the following steps:

1) Rewrite the systems of equations in three variables as systems of linear equations in two variables. Consider any two equations from the given set of three equations and eliminate one variable from those two equations.

Let us say we are eliminating the variable z. Then we will get an equation with the variables x and y and name this equation as (4). Consider another set of two equations and eliminate the same variable z. We will get another equation with the variables x and y and name this equation as (5).

2) Now, solve the two resulting equations (4) and (5) and find the value of x and y.

3) Substitute the value of x and y in any one of the three given equations and find the value of z.

4) So, the solution of the given system of equations is (x, y, z).

Let us solve an example to get clear idea of this process.

Example 1:

Solve the following systems of linear equations in three variables.

2x + 3y + 6z = 13

3x + 4y + 2z = 11

x - 3y + 5z = -5

Solution:

Step 1:

The given system contains three linear equations in three variables.

2x + 3y + 6z = 13 ----------------- (1)

3x + 4y + 2z = 11 ----------------- (2)

x - 3y + 5z = -5   ------------------ (3)

Step 2:

Consider the two equations (1) and (3).

Observe that the coefficients of y are 3 and -3 (additive inverses).

So, we can use elimination using addition method to eliminate the variable y.

     2x + 3y + 6z = 13 ----------------- (1)

(+)  x - 3y + 5z = -5   ----------------- (3)

------------------------------------------------

     3x         + 11z = 8                          

Now, we the resultant equation has only two variables, x and z. Name the equation as (4).

3x + 11z = 8 ----------------- (4)

Step 3:

Consider another set of two equations, (2) and (3).

Here, we need to eliminate the same variable y.

3x + 4y + 2z = 11 ----------------- (2)

 x - 3y + 5z = -5    ----------------- (3)

Use the elimination using multiplication to eliminate the variable y.

Multiply the equation (2) by 3 and multiply the equation (3) by 4 and add them.

      9x + 12y + 6z = 33

(+) 4x - 12y + 20z = -20

--------------------------------

      13x          + 26z = 13                    

We can simplify the above equation by dividing 13 on both the sides.

x + 2z = 1

So, the resultant equation also has only two variables, x and z. Name the equation as (5).

x + 2z = 1 -------------------- (5)

Step 4:

Now, solve the two equations (4) and (5) and find the value of x and z.

3x + 11z = 8 ----------------- (4)

x + 2z = 1 -------------------- (5)

Observe that the coefficient of x is 1 in the second equation.

So, we can use substitution method to solve these two linear equations.

Solve equation (5) for x. So, subtract 2z from both the sides and simplify to isolate x.

x + 2z - 2z = 1 - 2z

x = 1 - 2z

Now, substitute the value of x in equation (4).

3(1 - 2z) + 11z = 8

Multiply 3 and (1 - 2z) and simplify.

3 - 6z + 11z = 8

3 + 5z = 8

Subtract 3 from both the sides and simplify.

3 + 5z – 3 = 8 – 3

5z = 5

Divide by 5 on both the sides and simplify to isolate z.

z = 1

Now, replace the value of z by 1 in equation (5).

x + 2(1) = 1

x + 2 = 1

Subtract 2 from both the sides to isolate x and simplify.

x + 2 - 2 = 1 - 2

x = -1

So, we have found the value of the variables x and z as -1 and 1, respectively.

Step 5:

Substitute the value of x and z in the any one of the given three linear equations to find the value of y. Let us substitute in equation (3).

x - 3y + 5z = -5

Replace x by -1 and z by 1 in the above equation and simplify.

-1 - 3y + 5(1) = -5

-1 - 3y + 5 = -5

-3y + 4 = -5

Subtract 4 from each side of the equation.

-3y + 4 - 4 = -5 - 4

-3y = -9

Divide by -3 on both the sides.

y = 3

Step 6:

So, the solution of the given system of equations is (-1, 3, 1).

Practice Questions:

Solve the following systems of linear equations in three variables.

1) 2x + y – z = -1; -3x + 2y + 2z = 9; x + y – z = 0

Show Answer
x = -1; y = 2; z = 1

2) 6x + 3y + 2z = 2; -7x – 5y + 3z = 22; x + 2y – 5z = -24

Show Answer
x = 0; y = -2; z = 4

3) 3x + 5y – 7z = -1; -2x + 7y – 3z = -2; x – y + z = -1

Show Answer
x = -1; y = -1; z = -1

4) 5x + 3y – 2z = 0; 9x + 3y + 10z = 0; 7x – 9y + 3z = 0

Show Answer
x = 0; y = 0; z = 0

5) -2x + 4z = 9y + 38; z = 2x – 3y -19; 5x + 3y – 4z -1 = 0  

Show Answer
x = 5; y = -4; z = 3

Related Articles:

>> Systems of Linear Equations: Consistent, Inconsistent, Dependent, Independent, Number of Solutions

>> Solving Linear Equations: Graphing Method

>> Solving Linear Equations: Substitution Method

>> Elimination Method Using Addition and Subtraction

>> Elimination Method Using Multiplication

>> Linear Equations: Word Problems

Linear Equations Word Problems

Word problems for systems of linear equations are troublesome for most of the students in understanding the situations and bringing the word problem into equations. We tried to explain the trick of solving word problems for equations with two variables with an example.

Example:

2000 tickets were sold in an exhibition on Saturday. The cost of a ticket for an adult is $4 and for a child is $2. The total amount collected on Saturday was $6400. Find the number of adult tickets and child tickets sold on Saturday.

Solution:

Step 1:

Total number of tickets sold = 2000

Cost of ticket for an adult = $4

Cost of ticket for a child = $2

Total amount collected = $6400

We are asked to find the number of adult tickets and children tickets sold.

Let x be the number of adult tickets sold and y the number of child tickets sold on Saturday.

Step 2:

Total number of tickets sold = 2000

Number of adults tickets sold + Number of child tickets sold = 2000

x + y = 2000 --------------- (1)

Total amount collected = Amount collected from adult tickets + Amount collected from child tickets

Since there are x adults and the adult ticket cost is $4 per, the amount collected from adult tickets is 4x and since there are y children and the child ticket costs $2 per child, the amount collected from child tickets is 2y.

6400 = 4x + 2y

 It can be written as 4x + 2y = 6400 ---------------- (2)

Now, we got a system of two linear equations in two variables.

x + y = 2000 --------------- (1)

4x + 2y = 6400 ---------------- (2)

Step 3:

Solve the above two linear equations to find the value of x and y.

Multiply equation (1) with -2 and add the resulting equation and equation (2) to eliminate the variable y.

     -2x - 2y = -4000

(+) 4x + 2y = 6400

--------------------------------

       2x = 2400                                    

Divide by 2 on both the sides and simplify.

x = 1200

Step 4:

Now, substitute the value of x = 1200 in either of the two equations.

Let us substitute in equation (1). So, it becomes

1200 + y = 2000

Subtract 1200 from both sides of the equation and simplify.

1200 + y - 1200 = 2000 - 1200

y = 800

Step 5:

So, the solution for the given system of equations is (1200, 800), which means 1200 adult tickets and 800 child tickets were sold on Saturday. 

Note: The above problem can be also solved using substitution method since the coefficients of x and y in the first equation is 1.

 Practice Problems:

1. James bought 5 apples and 10 oranges for $4. Donald bought 3 apples and 9 oranges for $3. The shop keeper strictly told that there will not be any discounts. What is the cost of an apple and an orange?

Show Answer
Cost of an apple = $0.40
Cost of an orange = $0.20

2. Ana writes test to upgrade her level. The test has 25 questions for a total score of 150 points. Among the 25 questions, each multiple choice questions carries 3 points and the descriptive type questions carries 8 points. How many multiple choice questions and descriptive type questions are there in the test?

Show Answer
10 multiple choice and 15 descriptive

3. The sum of the digits of a two digit number is 7. When the digits are reversed, then number is decreased by 9. Find the number.

Show Answer
34 is a required number

4. The perimeter of a rectangle is 12 meters. The length is 3 more than twice its width. Find the dimensions of the rectangle.(This can also be solved just with one variable)

Show Answer
Length = 5m; Width = 1m

5. The sum of two numbers is 12. When three times the first number is added to 5 times the second number, the resultant number is 44. Find the two numbers.

Show Answer
The numbers are 8 and 4

Having problem in solving any of these questions? Post your doubt in comment. We will help you how to solve this.

Related Articles:

>> Systems of Linear Equations: Consistent, Inconsistent, Dependent, Independent, Number of Solutions

>> Solving Linear Equations: Graphing Method

>> Solving Linear Equations: Substitution Method

>> Elimination Method Using Addition and Subtraction

>> Elimination Method Using Multiplication

Elimination Method Using Multiplication

In our previous post, we have seen how to solve linear equations using elimination method by addition or subtraction where the coefficients of one of the variable is either same or additive inverses.
Suppose in the given system of equations, the coefficient of either variables are not additive inverse or not the same then we have to use elimination method using multiplication to solve the system of linear equations.

In this method, we eliminate one variable by making the coefficients same by multiplying the equations by a constant.

Let us explain this method with some examples.

Example 1:

Solve the system of equations by elimination method.

3x + 2y = 7

5x + 4y = 11

Solution:

Step 1:

Given equations are 3x + 2y = 7 ---- (1) and 5x + 4y = 11 ---- (2).

Here, the coefficients of x and y are neither same nor additive inverses.

So, multiply equation (1) by 2 to make the coefficients of y same.

Equation (1) becomes 6x + 4y = 14.

Step 2:

Now, subtract the equations.

     6x + 4y = 14

(-) 5x + 4y = 11

----------------------

               x = 3                                       

So, x = 3.

Step 3:

Substitute the value of x = 3 in either of the equations to find the value of y.

Let us substitute in equation (1).

3(3) + 2y = 7

9 + 2y = 7

Subtract 9 from both the sides and simplify.

9 + 2y - 9 = 7 - 9

2y = -2

Divide by 2 on both the sides to find the value of y.

y = -1

Step 4:

So, the solution for the given system of equations is (3, -1).

Note: In the above system of equations, we have multiplied one equation by a constant. In some cases, we need to multiply both the equations by a constant to make the coefficients same. Let us discuss this case in the next example.

Example 2:

Solve the system of equations using elimination method.

4x - 2y = -6

12x - 3y = -15

Solution:

Step 1:

Given equations are 4x - 2y = -6 ---- (1) and 12x - 3y = -15---- (2).

Here, the coefficients of x and y are not same or are not additive inverses.

So, we need to apply the elimination using multiplication method.

Step 2:

Multiply equation (1) by 3 on both the sides.

4x - 2y = -6 becomes 12x - 6y = -18.

And multiply equation (2) by -2 on both the sides.

12x - 3y = -15 becomes -24x - 6y = -30.

Now, observe that the coefficients of y in the two resultant equations, -6 and 6, are additive inverses.

Step 3:

Subtract the resultant two equations to find the value of x.

         12x - 6y = -18

(-) -24x - 6y = -30                     [-(24x - 6y = -30) = -24x + 6y = 30]

-----------------------------

-12x = 12                          

Divide both the equations by -12 to isolate x.

x = -1

Step 4:

Now, substitute the value of x = -1 in either equation (1) or (2).

Let us substitute in equation (1).

4(-1) - 2y = -6

-4 - 2y = -6

Add 4 on both sides of the equation and simplify.

-4 - 2y + 4 = -6 + 4

-2y = -2

Divide by -2 on both sides of the equation.

y = 1

Step 5:

So, the solution of the given system of equations is (-1, 1).

Note: There are many other combinations of multipliers that can be used to multiply the equations to make the coefficients same

We can use two methods to solve a system of linear equations by elimination method using multiplication.

Method 1:

Step 1: Eliminate y.

Step 2: Find the value of x.

Step 3: Substitute the value of x and find the value of y.

Method 2:

Step 1: Eliminate x.

Step 2: Find the value of y.

Step 3: Substitute the value of y and find the value of x.

We have used method 1 in the above problem. We can also use method 2. Try to solve the above problem using the method 2.

The following table will guide us to choose the best method to solve.

Methods	When to choose
Graphing	When we are asked to estimate the solution. Graphing sometimes does not give the exact solution.
Substitution	If the coefficient of one of the variables in either equation is 1 or -1.
Elimination using addition	If one of the variables in the given two equations has the opposite coefficients (additive inverses).
Elimination using subtraction	If one of the variables in the given two equations has the same coefficient.
Elimination using multiplication	If none of the variables has the coefficients 1 or -1 and if the variables cannot be eliminated by using addition or subtraction.

Let us solve one more example on solving system of equations.

Example 3:

Choose the best method to solve the given system of equation and find the solution.

5x - 3y = 12

7x - 9y = 15

Solution:

Step 1:

Here, we need to find the exact solution of the given system. The algebraic method is the best way to find the exact solution.

None of the variables has coefficients 1 or -1. So, we cannot use substitution method.

The coefficients of x and y are neither same nor additive inverses. So, we cannot use the elimination using addition or subtraction. Hence, elimination using multiplication is the best method to solve the given system of equations.

Step 2:

Multiply the first equation by -3.

-15x + 9y = -36

The second equation is 7x - 9y = 15.

Now, we can observe that the coefficients of y in the two equations are additive inverses.

Add the above two equations to eliminate y.

     -15x + 9y = -36

(+)    7x - 9y = 15

--------------------------

       -8x = -21                                       

Divide by -8 on both the sides to isolate x.

x = ²¹/₈

Step 3:

Substitute the value of x = ²¹/₈ in the first equation 5x - 3y = 12 to find the value of y.

5(²¹/₈) – 3y = 12

¹⁰⁵/₈ – 3y = 12

Subtract ¹⁰⁵/₈ from both the sides of the equation and simplify.

(¹⁰⁵/₈) - 3y -(¹⁰⁵/₈)  = 12 - (¹⁰⁵/₈)

-3y = 12 - (¹⁰⁵/₈)

-3y = ^{(96 – 105)}/₈

-3y = ^-9/₈

Divide by -3 on both the sides and simplify.

y = ³/₈

Step 4:

So, the solution is (²¹/₈, ³/₈).

Practice Questions:
Solve the given system of equations using elimination method

1) 2x + 3y = 4; 3x + 2y = 1

Show Answer
x = -1 and y = 2

2)  5x – y = -14; 2x + 4y = 12

Show Answer
x = -2 and y = 4

3) -7x + 8y = -10; 4x – 3y = 12

Show Answer
x = 6 and y = 4

4) 6x + 2y = -24; x + 7y = 16

Show Answer
x = -5 and y = 3

5) -7x + 10y = 20; 8x – 5y = 35

Show Answer
x = 10 and y = 9

Related Articles:

>> Systems of Linear Equations: Consistent, Inconsistent, Dependent, Independent, Number of Solutions

>> Solving Linear Equations: Graphing Method

>> Solving Linear Equations: Substitution Method

>> Elimination Method Using Addition and Subtraction

You may also interested in this:

>> Equations in single variable: Quadrilaterals

>> Equations in single variable: Algebra

>> Age Word Problems

Elimination Method: Linear Equations (elimination method using addition and subtraction)

Elimination method is one of the best methods of solving the linear equations. In this method, we eliminate one variable from the equations and then find the value of the variable. Then substitute the value of the variable in either original equation to find the value of the eliminated variable.

Solving the system of linear equations by elimination method can be done in three ways. They are:-

a) Elimination using addition.

b) Elimination using subtraction.

c) Elimination using multiplication.

In this post we are going to discuss elimination using addition and subtraction. Elimination method using multiplication will be discussed in next post.

Elimination Method Using Addition:

In this method, we eliminate one like variable by adding the two equations and solve the resulting equation for the other variable and substitute the value of this variable in either of the given equations to find the value of the eliminated variable.

Elimination method of addition is easy to use for the equations with at least one like variable whose co-efficient are additive inverse of each other.

Let us explain this method with an example.

Example:

Solve the system of equations by elimination method. 

4x - 2y = 14

6x + 2y = 16

Solution:

Step 1:

The given system of equations is

4x – 2y = 14

6x + 2y = 16

 Observe that the coefficients of y terms are -2 and 2 which are additive inverses. So, by adding the above two equations, we can eliminate the y variable and find the value of the variable x.

Step 2:

Write the two equations one below the other as shown and add.

      4x - 2y = 14

(+) 6x + 2y = 16            

----------------------

       10x      = 30                      

Now, the variable y is eliminated.

Divide by 10 on both the sides to find the value of x.

So, x = 3.

Step 3:

Now, substitute the value of x = 3 in either of the given equations to find the value of y. Let us substitute in the first equation 4x - 2y = 14 and simplify.

4(3) - 2y = 14

12 - 2y = 14

Subtract 12 from both the sides and simplify.

12 - 2y - 12 = 14 - 12

-2y = 2

Divide both the sides by -2 to isolate y.

y = -1

Step 4:

So, the solution of the given system of linear equations is (3, -1).

Elimination Method Using Subtraction:

In this method, we eliminate one variable (with the same coefficient) by subtracting the two equations to find the value of other variable and then substitute the value of this variable in either of the given equations to find the value of the eliminated variable.

Let us explain this method with an example.

Example:

Solve the system of equations by elimination method. 

5m + 6n = 7

5m + 2n = -1

Solution:

Step 1:

Observe that the co-efficient of m terms are the same. 

Subtract the given equations to eliminate the m terms and find the value of the variable n.

Step 2:

Write the two equations one below the other as shown and subtract.

     5m + 6n = 7

(-) 5m + 2n = -1                     [-(5m + 2n = -1) = -5m - 2n = 1]

---------------------

               4n = 8                                   

Now, the variable m is eliminated.

Divide both the sides by 4 to find the value of n.

n = 2

Step 3:

Now, substitute the value of n = 2 in either of the equations to find the value of m.

Let us substitute in the second equation 5m + 2n = -1 and simplify.

5m + 2(2) = -1

5m + 4 = -1

Subtract 4 from both sides of the equation and simplify.

5m + 4 - 4 = -1 - 4

5m = -5

Divide by 5 on both the sides to isolate m.

m = -1

Step 4:

So, (-1, 2) is a solution for the given system of equations.

Practice Questions:

Solve the following linear equations using Elimination method:

a) 2x – 7y = 12; -3x + 7y = -11

Show Answer
x = -1 and y = -2

b) 6x + 5y = -5; -6x – 7y = -5

Show Answer
x = -5 and y = 5

c) 3x + 4y = 1; 3x -2y = -5

Show Answer
x = -1 and y = 1

d) 2x – 5y = -2; -x – 5y = -14

Show Answer
x = 4 and y = 2

e)  7x – 8y = 19; 7x + 8y = 51

Show Answer
x = 5 and y = 2

Related Articles:

>> Systems of Linear Equations: Consistent, Inconsistent, Dependent, Independent, Number of Solutions

>> Solving Linear Equations: Graphing Method

>> Solving Linear Equations: Substitution Method

You may also interested in this:

>> Venn diagram: Word problems

>> Percent Word problems

>> Equations in single variable: Quadrilaterals

>> Equations in single variable: Algebra

>> Age Word Problems

Substitution Method: Solve Linear Equations

Substitution method is an algebraic method of solving the system of linear equations. Using this method, we can find the exact solution for the equations.

Let us explain this method using few examples.

Example 1:

Solve the system of linear equations using substitution method.

y = 2x

x + 2y = 5

Solution:

Step 1:

Given equations are y = 2x and x + 2y = 5.

The value of y is directly given in the first equation. So, substitute the value of y = 2x in the second equation x + 2y = 5 and simplify.

So, x + 2y = 5 becomes x + 2(2x) = 5.

x + 4x = 5

Add 4x and x.

5x = 5

Divide by 5 on both the sides to isolate x.

x = 1

Step 2:

Substitute the value x = 1 to find the value of y.

Replace the value of x by 1 in the equation y = 2x.

y = 2 (1)

y = 2

Step 3:

So, the solution of the given system of equations is (1, 2).

Example 2:

Using substitution method, solve the given linear equations.

x – y = 3

3x + 2y = 9

Solution:

 Step 1:

Given equations are x - y = 3 and 3x + 2y = 9.

We need to find the value of variables by substituting the value of one variable in the other equation.

Let us find the value of x from the first equation.

To find the value of x, solve for x.

Add y on both the sides of the equation x - y = 3 to find the value of x and then simplify.

x - y + y = 3 + y

x = y + 3

Step 2:

Replace the value of x by y + 3 in the second equation 3x + 2y = 9 and simplify it.

3(y + 3) + 2y = 9

3y + 9 + 2y = 9

Combine the like terms.

(3y + 2y) + 9 = 9

5y + 9 = 9

Subtract 9 from both the sides.

5y + 9 - 9 = 9 - 9

5y = 0

Divide by 5 on both the sides.

y = 0

Step 3:

Now, substitute the value of y = 0 in any of the equation whichever is easy to solve.

So, substitute the value of y in the first equation x - y = 3.

x - 0 = 3

x = 3

Step 4:

So, the solution is (3, 0).

Example 3:

Use substitution method to solve the given system of linear equations.

 2x + y = 4

4x + 2y = 8

Solution:

Step 1:

Given equations are 2x + y = 4 and 4x + 2y = 8.

We need to find the value of variables by substituting the value of one variable in the other equation.

Let us find the value of y from the first equation 2x + y = 4.

Subtract 2x from both the sides,

2x - 2x + y = 4 - 2x

y = 4 - 2x

Step 2:

Now, substitute the value of y = 4 - 2x in the second equation 4x + 2y = 8.

4x + 2(4 - 2x) = 8

4x + 8 - 4x = 8

8 = 8

Step 3:

We end up with a statement 8 = 8. This is a true statement which means that the given system of equations has infinitely many solutions.

Important note:

While solving the system of equations, we may arrive at some statements like 5 = 5, -5 = -5, and 3 = 4. 

If we arrive at a true statement (E.g. 4 = 4), then the system has infinitely many solutions and the graph of both the equations are the same.

If we arrive at a false statement (E.g. 4 = 3), then the system has no solution and the lines are parallel.

Practice questions:

Use substitution method to solve the given system of equations.

a) -2x + 9y = -9; 9x – 2y = 2

Show Answer
x = 0 and y = -1

b) x + 2y = -3; -2x + 4y = 10

Show Answer
x = -4 and y = 1/2

c) x + y = 3; x + y = 1

Show Answer
Lines are parallel. It has no solution

d) x + 5y = 8; 2x – 3y = -10

Show Answer
x = -2 and y = 2

e) 7x – y = 10; 9x + 2y = 3

Show Answer
x = 1 and y = -3

Related Articles:

>> Systems of Linear Equations: Consistent, Inconsistent, Dependent, Independent, Number of Solutions

>> Solving Linear Equations: Graphing Method

You may also interested in this:

>> Venn diagram: Word problems

>> Equations in single variable: Quadrilaterals

>> Equations in single variable: Algebra

>> Age Word Problems