Linear Equations with Two Variables

Solving equations is probably the most important and interesting topic in algebra. There are a wide variety of equations. In this article, we will learn how to solve equations, but more specifically, we will look at solving equations with two variables. The use of exercises with answers will make the concepts easier to understand.

ALGEBRA

Relevant for…

Learning to solve linear equations with two variables.

See methods

Contents

ALGEBRA

Relevant for…

Learning to solve linear equations with two variables.

See methods

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 $Ax+By+C=0$ , where the coefficients $A$ and $B$ are different from 0.

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

At first glance, the equation $4{{x}^2}+6x-2y-4{{x}^2}=5$ 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 ${{x}^2}$ are canceled and we have the equation $6x-2y=5$ . 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:

• $x+3y+5=12$

• $\frac{1}{3}x+3y=\frac{3}{2}+4x$

• $2{{x}^{2}}+4x-3y=2{{x}^{2}}+8$

• $2x+{{x}^{3}}-{{x}^{3}}=9-5y$

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 $2x+4y=3x-20$ .

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

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

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.

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.

3. Isolate the variable:

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

4. Solve for the variable using multiplication or division.

5. Verify the answer.

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Linear equations with two variables – Examples with answers

EXAMPLE 1

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

Solution:

1. Substitute the value of the variable:

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

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

2. Simplify:

There are no parentheses.

There are no fractions.

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

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

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

$2x=9-x$

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

$3x=9$

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

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

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

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

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

$6+8=10-3+7$

$14=14$

This is true

Answer: $y=2, x=3$

EXAMPLE 2

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

Solution:

1. Substitute the value of the variable:

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

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

2. Simplify:

There are no parentheses.

There are no fractions.

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

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

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

$-2y=-10$

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

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

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

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

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

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

$15=15$

This is true

Answer: $y=5, x=4$

EXAMPLE 3

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

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 $x =1$ , then we would have:

$3x-2y=20+x$

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

$3-2y=21$

$-2y=21-3$

$-2y=18$

$y=-9$

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

$3x-2y=20+x$

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

$6-2y=22$

$-2y=22-6$

$-2y=16$

$y=-8$

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

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