Linear equations with two unknowns are equations that have two variables that are not raised to any power. An example of an equation with two unknowns is *x*+2*y*=5. These equations can only be solved if we know the value of one of the variables. Otherwise, the equation has an infinite number of solutions.

In this article, we will become familiar with solving linear equations in two unknowns with worked examples to help understand the concepts.

## What are equations with two variables?

Linear equations have the characteristic that they contain only variables raised to the first power after being reduced to their simplest form. Equations with two unknowns or variables can be written in the general form $latex Ax+By+C=0$, where the coefficients $latex A$ and $latex B$ are different from 0.

For example, the equations $latex 4x+3y=10$ and $latex 5x-5y=10$ are linear equations with two variables.

At first glance, the equation $latex 4{{x}^2}+6x-2y-4{{x}^2}=5$ appears not to be linear. However, when written in its simplest form, that is, when we combine like terms and simplify, the terms that contain $latex {{x}^2}$ are canceled and we have the equation $latex 6x-2y=5$. Thus, it is a linear equation with two variables.

Generally, a linear equation with two variables has an infinite number of solutions. However, if we have a defined value for one of the variables, the equation will have only one solution.

A system of equations with two variables can have only one solution, infinite solutions, or no solution depending on the values of its coefficients.

### EXAMPLES

The following equations are linear equations with two variables:

• $latex x+3y+5=12$

• $latex \frac{1}{3}x+3y=\frac{3}{2}+4x$

• $latex 2{{x}^{2}}+4x-3y=2{{x}^{2}}+8$

• $latex 2x+{{x}^{3}}-{{x}^{3}}=9-5y$

## Rule of equations

Any operation (addition, subtraction, multiplication, division) can be performed on one side of the equation, as long as the same operation is performed on the opposite side.

For example, suppose we have the equation $latex 2x+4y=3x-20$.

If we want to, we can subtract 6 from both sides of the equation: $latex 2x+4y-6=3x-20-6$.

In this case, it doesn’t help much to have 6 subtracted from both sides of the equation. However, we can simplify the equations and eventually solve them if we apply the correct operations to both sides.

Now let’s look at a method with which we can solve equations with two variables.

## Method for solving linear equations with two variables

**Step 1.** If we have a value for one of the variables, we plug that value into the equation. If we don’t have the value of any variable, the equation automatically has an infinite number of solutions.

**Step 2.** Simplify both sides:

a. Remove the parentheses (using the distributive property) if there are any.

b. Eliminate the fractions (multiply by the least common multiple).

c. Simplify the like terms.

**Step 3.** Isolate the variable:

We perform operations to move the variable to one side of the equation and the constants to the other side.

**Step 4.** Solve for the variable using multiplication or division.

**Step 5.** Verify the answer.

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## Linear equations with two variables – Examples with answers

### EXAMPLE 1

- Given the value $latex y=2$, find the value of$latex x$ in the equation $latex 2x+4y=10-x+7$.

**Solution:**

1. Substitute the value of the variable:

$latex 2x+4(2)=10-x+7$

$latex 2x+8=10-x+7$

2. Simplify:

There are no parentheses.

There are no fractions.

Combine like terms: $latex 2x+8=17-x$.

3. Isolate the variable: move the 8 to the right and the -x to the left:

$latex 2x+8-8=17-x-8$

$latex 2x=9-x$

$latex 2x+x=9-x+x$

$latex 3x=9$

4. Isolate completely the $latex x$: divide both sides by 3:

$latex \frac{3}{3}x=\frac{{9}}{3}$

$latex x=\frac{{9}}{3}=3$

5. Verify the answer: replace the values of the variables in the original equation:

$latex 2(3)+4(2)=10-(3)+7$

$latex 6+8=10-3+7$

$latex 14=14$

This is true

**Answer:** $latex y=2, x=3$

### EXAMPLE 2

- If we have $latex x=4$, find the value of $latex y$ in the equation $latex 5x-2y+5=12-3x+15$.

**Solution:**

1. Substitute the value of the variable:

$latex 5(4)-2y+5=12-3(4)+15$

$latex 20-2y+5=12-12+15$

2. Simplify:

There are no parentheses.

There are no fractions.

Combine like terms: $latex 25-2y=15$.

3. Isolate the variable: move the 25 to the right:

$latex 25-2y-25=15-25$

$latex -2y=-10$

4. Solve for $latex y$: divide both sides by -2:

$latex \frac{-2}{-2}y=\frac{{-10}}{-2}$

$latex y=\frac{{-10}}{-2}=5$

5. Verify the solution: substitute the values of the variables in the original equation:

$latex 5(4)-2(5)+5=12-3(4)+15$

$latex 20-10+5=12-12+15$

$latex 15=15$

This is true

**Answer:** $latex y=5, x=4$

### EXAMPLE 3

- Find the value of $latex x$ and $latex y$ in the equation $latex 3x-2y=20+x$.

**Solution:**

We do not have any value of any of the variables, so this equation has an infinite number of solutions. For example, suppose that $latex x =1 $, then we would have:

$latex 3x-2y=20+x$

$latex 3(1)-2y=20+1$

$latex 3-2y=21$

$latex -2y=21-3$

$latex -2y=18$

$latex y=-9$

Now, suppose we have $latex x=2$, then we would have:

$latex 3x-2y=20+x$

$latex 3(2)-2y=20+2$

$latex 6-2y=22$

$latex -2y=22-6$

$latex -2y=16$

$latex y=-8$

We could go through the process with different values for $latex x$ and we would always get different answers. This means that there are an infinite number of solutions.

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### EXAMPLE 4

- If we have $latex x=6$, find the value of $latex y$ in the equation $latex x+4(y+2)=10+2y+2x$.

**Solution:**

1. Substitute the value of the variable:

$latex 6+4(y+2)=10+2y+2(6)$

$latex 6+4(y+2)=10+2y+12$

2. Simplify:

Remove parentheses: $latex 6+4y+8=10+2y+12$.

There are no fractions.

Combine like terms: $latex 4y+14=2y+22$.

3. Isolate the variable: move the 14 to the right and 2y to the left:

$latex 4y+14-14=2y+22-14$

$latex 4y=2y+8$

$latex 4y-2y=2y+8-2y$

$latex 2y=8$

4. Solve for $latex y$: divide both sides by 2:

$latex \frac{2}{2}y=\frac{{8}}{2}$

$latex y=\frac{{8}}{2}=4$

5. Verify the solution: substitute the values of the variables in the original equation

$latex 6+4(4+2)=10+2(4)+2(6)$

$latex 6+4(6)=10+8+12$

$latex 30=30$

This is true

**Answer:** $latex y=4, x=6$

### EXAMPLE 5

- Given the value $latex y=2$, find the value of $latex x$ in the equation $latex \frac{1}{3}x+2y+\frac{1}{2}x=y+x+1$.

**Solution:**

1. Substitute the value of the variable:

$latex \frac{1}{3}x+2(2)+\frac{1}{2}x=2+x+1$

$latex \frac{1}{3}x+4+\frac{1}{2}x=2+x+1$

2. Simplify:

There are no parentheses.

Simplify fractions: multiply by 6: $latex 2x+24+3x=12+6x+6$.

Combine like terms: $latex 5x+24=18+6x$.

3. Isolate the variable: move the 24 to the right and the 6x to the left:

$latex 5x+24-24=18+6x-24$

$latex 5x=6x-6$

$latex 5x-6x=6x-6-6x$

$latex -x=-6$

4. Solve for $latex x$: divide both sides by -1:

$latex \frac{-1}{-1}x=\frac{{-6}}{-1}$

$latex x=\frac{{-6}}{-1}=6$

5. Verify the solution: substitute the values of the variables in te original equation:

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

$latex 2+4+3=2+6+1$

$latex 9=9$

This is true

**Answer:** $latex y=2, x=6$

## See also

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