Linear Equations with Two Unknowns – Examples and Practice Problems

We can solve linear equations with two unknowns if we know the value of one of the variables. Otherwise, if we do not know the value of one variable, we can conclude that the equation has an infinite number of solutions.

In this article, we will look at a summary of the process used to solve linear equations examples with two unknowns. In addition, we will look at several examples with answers to master the process.

ALGEBRA
linear equations with two variables

Relevant for

Solving linear equation examples with two unknowns.

See examples

ALGEBRA
linear equations with two variables

Relevant for

Solving linear equation examples with two unknowns.

See examples

Summary of linear equations with two unknowns

For an equation to be of the first degree, all its variables must have a maximum power of 1. In this case, the two unknowns must have a power of 1. For example, the equations 3x+2y=5 and 3y=2x-4 are first degree equations with two unknowns. To solve these types of equations, we can follow the following steps:

Step 1: Substitute the known value of one variable. If we don’t have the value of a variable, the equation automatically has infinite solutions.

Step 2: Simplify: We remove grouping signs (like parentheses), remove fractions, and simplify like terms.

Step 3: Solve for the variable. We move all the variables to one side of the equation and the constants to the other side.

Step 4: We solve using division or multiplication.


Examples with answers of linear equations with two unknowns

With the following examples, you can practice solving linear equations with two unknowns. Each example shows the respective process used to find the solution. Try to solve the problems first before looking at the solution.

EXAMPLE 1

If the value of y is equal to 5, find the value of x in the equation 3x-4y=10.

Step 1: Substitute: We have that y=5, therefore:

3x-4y=10

3x-4(5)=10

3x-20=10

Step 2: Simplify: We have nothing to simplify:

Step 3: Solve for the variable: We add 20 to both sides:

3x-20=10

3x-20+20=10+20

3x=30

Step 4: Solve: We divide both sides by 3:

 \frac{3x}{3}=\frac{30}{3}

x=10

EXAMPLE 2

We have that the value of y is equal to -3. Solve the equation -3x+5y=-6 for x.

Step 1: Substitute: We have that y=-3, therefore:

-3x+5y=-6

-3x+5(-3)=-6

-3x-15=-6

Step 2: Simplify: We do not have like terms.

Step 3: Solve for the variable: We add 15 to both sides:

-3x-15+15=-6+15

-3x=9

Step 4: Solve: We divide both sides by -3:

 \frac{-3x}{-3}=\frac{9}{-3}

x=-3

EXAMPLE 3

If the value of x is equal to -2, solve the equation 4y+2(2y+3)=3x-4 for y.

Step 1: Substitute: We have that x=-2, therefore:

4y+2(2y+3)=3(-2)-4

4y+2(2y+3)=-6-4

Step 2: Simplify: We expand the parentheses and combine like terms:

4y+2(2y+3)=-6-4

4y+4y+6=-10

8y+6=-10

Step 3: Solve for the variable: We subtract 6 from both sides:

8y+6-6=-10-6

8y=-16

Step 4: Solve: We divide both sides by 8:

 \frac{8y}{8}=\frac{16}{-8}

y=-2

EXAMPLE 4

We have that the value of x is equal to 5. Solve the equation 3x+2(-4x+5)=3y+6 for y.

Step 1: Substitute: We substitute x=5 in the equation:

3x+2(-4x+5)=3y+6

3(5)+2(-4(5)+5)=3y+6

15+2(-20+5)=3y+6

Step 2: Simplify: We expand the parentheses and combine like terms:

15+2(-20+5)=3y+6

15+2(-15)=3y+6

15-30=3y+6

-15=3y+6

Step 3: Solve for the variable: We subtract 6 from both sides:

-15-6=3y+6-6

-21=3y

Step 4: Solve: We divide both sides by 3:

 \frac{-21}{3}=\frac{3y}{3}

-7=y

EXAMPLE 5

If the value of z equals 5, solve the equation 4y+2z=2(3y+10)+z-11 for y.

Step 1: Substitute: We substitute z=5 in the equation:

4y+2(5)=2(3y+10)+5-11

4y+10=2(3y+10)+5-11

Step 2: Simplify: We expand the parentheses and combine like terms:

4y+10=6y+20+5-11

4y+10=6y+14

Step 3: Solve for the variable: We subtract 10 and 6 y from both sides:

4y+10-10=6y+14-10

4y=6y+4

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

-2y=4

Step 4: Solve: We divide both sides by -2:

 \frac{-2y}{-2}=\frac{4}{-2}

y=-2

EXAMPLE 6

If the value of y is equal to -3, solve the equation \frac{y+1}{2}+2x=2(2y+6)+x+2 for x.

Step 1: Substitute: We have that y=-3, therefore:

\frac{-3+1}{2}+2x=2(2(-3)+6)+x+2

\frac{-3+1}{2}+2x=2(-6+6)+x+2

Step 2: Simplify: We combine like terms and simplify:

\frac{-2}{2}+2x=2(0)+x+2

-1+2x=x+2

Step 3: Solve for the variable: We add 1 and subtract x from both sides:

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

2x=x+3

2x-x=x+3-x

x=3

Step 4: Solve: We have already found the solution:

 x=3

EXAMPLE 7

Solve the equation 2x+2y=3x+10 for y.

Step 1: Substitute: In this case we do not have any given value, so automatically, the equation has an infinite number of solutions. For example, suppose we have x = 0, then we would have:

2x+2y=3x+10

2(0)+2y=3(0)+10

2y=10

y=5

If we now have x = 1, we have:

2x+2y=3x+10

2(1)+2y=3(1)+10

2+2y=3+10

2+2y=13

2y=15

y=15/2

We could continue with different values and each time we would get different results, so by not having a specified value of a variable, the equation has infinite solutions.


Linear equations with two unknowns – Practice problems

Use the following practice problems to practice solving linear equations with two unknowns. Simply choose an answer and verify it by clicking “Check”. You can look at the solved examples above carefully if you have trouble solving these exercises.

If we have y=-2, what is the value of x in 2x+4y=-4?.

Choose an answer






Solve the equation 4x+2y=2x-2 if we have y=5.

Choose an answer






We have z=10. Find the value of x in the equation 4x+3(z-10)=z+10.

Choose an answer






Solve the equation 2x-2y+5=x+3.

Choose an answer






Solve the equation \frac{x+1}{3}+5=\frac{y+2}{2}+x+4 if we have y=-2.

Choose an answer







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