Solving a system of three linear equations

 
 
Linear systems in three unknowns.jpeg
 
 
 

Number of solutions to the linear system

In this lesson we’ll look at how to solve systems of three linear equations in three variables.

If a system of three linear equations has solutions, each solution will consist of one value for each variable.

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If the three equations in such a linear system are “independent of one another,” the system will have either one solution or no solutions. All the systems of three linear equations that you’ll encounter in this lesson have at most one solution.

 
 

How to solve systems of linear equations, when the system is made of three equations


 
 

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Solve the system of equations

Solve the system of equations.

[1]  x5y+z=17-x-5y+z=17

[2]  5x5y+5z=5-5x-5y+5z=5

[3]  2x+5y3z=102x+5y-3z=-10

Notice that the coefficients of yy in equations [1] and [3] are 5-5 and 55, respectively. If we add these two equations, the yy terms will cancel (we’ll eliminate the variable yy) and we’ll get an equation in only the variables xx and zz.

(x5y+z)+(2x+5y3z)=17+(10)(-x-5y+z)+(2x+5y-3z)=17+(-10)

Remove parentheses and combine like terms.

x5y+z+2x+5y3z=1710-x-5y+z+2x+5y-3z=17-10

x+2x5y+5y+z3z=1710-x+2x-5y+5y+z-3z=17-10

x2z=7x-2z=7

You might have also noticed that the coefficients of yy in equations [2] and [3] are 5-5 and 55, respectively, so we can add these two equations to get another equation in only the variables xx and zz.

(5x5y+5z)+(2x+5y3z)=(5)+(10)(-5x-5y+5z)+(2x+5y-3z)=(5)+(-10)

Remove parentheses and combine like terms.

5x5y+5z+2x+5y3z=510-5x-5y+5z+2x+5y-3z=5-10

5x+2x5y+5y+5z3z=510-5x+2x-5y+5y+5z-3z=5-10

3x+2z=5-3x+2z=-5

The coefficients of zz in our two new equations are 2-2 and 22, respectively. If we add these two equations, we can eliminate the variable zz, and then solve for xx.

x2z=7x-2z=7

3x+2z=5-3x+2z=-5

(x2z)+(3x+2z)=7+(5)(x-2z)+(-3x+2z)=7+(-5)

Remove parentheses and combine like terms.

x2z3x+2z=75x-2z-3x+2z=7-5

x3x2z+2z=75x-3x-2z+2z=7-5

2x=2-2x=2

x=1x=-1

Choose one of the new equations, and plug in 1-1 for xx, and then solve for zz. We’ll choose x2z=7x-2z=7.

12z=7-1-2z=7

2z=8-2z=8

z=4z=-4

Now choose one of the three original equations, and plug in 1-1 for xx and 4-4 for zz, and then solve for yy. We’ll choose equation [1].

[1]  x5y+z=17-x-5y+z=17

(1)5y+(4)=17-(-1)-5y+(-4)=17

Simplify and solve for yy.

15y4=171-5y-4=17

5y+14=17-5y+1-4=17

5y3=17-5y-3=17

5y=20-5y=20

y=4y=-4

The solution is (1,4,4)(-1,-4,-4) or x=1x=-1, y=4y=-4, and z=4z=-4.


Let’s do one more.


If a system of three linear equations has solutions, each solution will consist of one value for each variable.

Example

Use any method to solve the system of equations.

[1]  3a3b+4c=233a-3b+4c=-23

[2]  a+2b3c=25a+2b-3c=25

[3]  4ab+c=254a-b+c=25

None of the terms with the same variable have the same coefficient (or coefficients that are equal in absolute value but opposite in sign). So we’ll need to multiply one of the equations by some number such that by combining the resulting equation with one of the other two equations, we’ll be able to eliminate a variable. Let’s multiply equation [2] by 33, so we can eliminate the variable aa by subtracting the resulting equation from equation [1].

3(a+2b3c)=3(25)3(a+2b-3c)=3(25)

Remove the parentheses.

[4]  3a+6b9c=753a+6b-9c=75

Now let’s subtract equation [4] from equation [1], which will give us an equation in only the variables bb and cc.

[1]  3a3b+4c=233a-3b+4c=-23

[4]  3a+6b9c=753a+6b-9c=75

(3a3b+4c)(3a+6b9c)=(23)(75)(3a-3b+4c)-(3a+6b-9c)=(-23)-(75)

Eliminate the parentheses, and then combine like terms.

3a3b+4c3a6b+9c=23753a-3b+4c-3a-6b+9c=-23-75

3a3a3b6b+4c+9c=23753a-3a-3b-6b+4c+9c=-23-75

[5]  9b+13c=98-9b+13c=-98

We need to get another equation in only the variables bb and cc. Let’s use equations [2] and [3].

This time we need to multiply equation [2] by 44, so we can subtract it from equation [3] and eliminate the variable aa.

[2]  a+2b3c=25a+2b-3c=25

Remove the parentheses.

4(a+2b3c)=4(25)4(a+2b-3c)=4(25)

[6]  4a+8b12c=1004a+8b-12c=100

Now we’ll subtract equation [6] from equation [3].

[3]  4ab+c=254a-b+c=25

(4ab+c)(4a+8b12c)=(25)(100)(4a-b+c)-(4a+8b-12c)=(25)-(100)

Eliminate the parentheses, and then combine like terms.

4ab+c4a8b+12c=251004a-b+c-4a-8b+12c=25-100

4a4ab8b+c+12c=251004a-4a-b-8b+c+12c=25-100

[7]  9b+13c=75-9b+13c=-75

With [5] and [7], we now have a system of two equations in the variables bb and cc.

[5]  9b+13c=98-9b+13c=-98

[7]  9b+13c=75-9b+13c=-75

If we subtract [7] from [5], we get

(9b+13c)(9b+13c)=98(75)(-9b+13c)-(-9b+13c)=-98-(-75)

Eliminate the parentheses, and combine like terms.

9b+13c+9b13c=98+75-9b+13c+9b-13c=-98+75

0=230=-23

Isn’t that impossible?

Actually, it isn’t impossible, but if something like that happens, it means that the original system of three equations has no solution.

 
 

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