Linear Equations with Two Variables - Examples, Exercises and Solutions

Question Types:
Linear Equations with Two Variables: Using fractionsLinear Equations with Two Variables: false statements and infinite solutionsLinear Equations with Two Variables: Applying the formula

An equation that has two variables: X X and Y Y .
y=a×x+b y=a\times x+b
To solve a linear equation that has two variables, we must find a pair of values for X X and for Y Y that preserve the equation.
How will we do it?

  1. Try to isolate one variable, whichever you prefer, then leave it alone on one side so that it does not have a value by itself.
  2. Place any number you want instead of the variable you have not isolated and discover the value of the isolated variable.

In this way, you will be able to discover the pair of variables that satisfy the equation in question.

This type of equations generally has infinite solutions.
If you create a value table for this equation and treat it as a function, you can plot it on the Cartesian plane and see what it looks like graphically.

Mathematical representation of a linear equation in two variables: y = ax + b. A foundational concept in algebra, demonstrating the slope-intercept form, where 'a' represents the slope and 'b' the y-intercept. Featured in a guide on solving linear equations with two variables.

Detailed explanation

Practice Linear Equations with Two Variables

Question 1

Solve the following system of equations:

\( \begin{cases} x+y=8 \\ x=5-y \end{cases} \)

Question 2

\( x+y=8 \)

\( x-y=6 \)

Question 3

\( 3x-y=5 \)

\( 5x+2y=12 \)

Question 4

\( 6x+4y=18 \)

\( -2x+3y=20 \)

Question 5

\( 6x+y=12 \)

\( 3y+2x=20 \)

Examples with solutions for Linear Equations with Two Variables

Exercise #1

Solve the following system of equations:

{x+y=8x=5−y \begin{cases} x+y=8 \\ x=5-y \end{cases}

Video Solution

Step-by-Step Solution

Note that in the current system of equations, one of the variables is isolated alone on the left side of the equation:

{x‾+y=8x=5−y‾ \begin{cases} \underline{x}+y=8 \\ \bm{x=\underline{5-y}} \\ \end{cases}

Therefore, we can use the substitution method and substitute the entire expression that x equals in the second equation in place of x in the first equation (marked with an underline in both equations above) and thus we'll get one equation with one variable:

5−y‾+y=8 \underline{ 5-y}+y=8

where we highlight the equation in which the variable we substituted is isolated in order to return to it later after we find the value of y from solving the equation we got, and this is to find using it the value of x corresponding to that y value we found, therefore we highlighted this equation above.

From here - we'll continue and solve the single-variable equation we got,

First- we'll combine like terms on the left side of the resulting equation:

5−y+y=85=8 5-y+y=8 \\ 5=8

We'll stop here and notice that y cancelled out in the current equation and we got a false statement, this is because clearly:

5≠8 5\neq8 meaning-

We got a false statement regardless of the variables' values,

We can conclude from here that the system of equations has no solution, because no matter which values we substitute for the variables - we won't get a true statement in both equations together.

Therefore the correct answer is answer D.

Answer

There is no solution.

Exercise #2

x+y=8 x+y=8

x−y=6 x-y=6

Video Solution

Answer

x=7,y=1 x=7,y=1

Exercise #3

3x−y=5 3x-y=5

5x+2y=12 5x+2y=12

Video Solution

Answer

x=2,y=1 x=2,y=1

Exercise #4

6x+4y=18 6x+4y=18

−2x+3y=20 -2x+3y=20

Video Solution

Answer

x=−1,y=6 x=-1,y=6

Exercise #5

6x+y=12 6x+y=12

3y+2x=20 3y+2x=20

Video Solution

Answer

x=1,y=6 x=1,y=6

Question 1

\( -x+y=14 \)

\( 5x+2y=7 \)

Question 2

\( 4x+3y=-11 \)

\( 3x-2y=-4 \)

Question 3

\( 4x-8y=16 \)

\( -x-2y=24 \)

Question 4

\( 6x-2y=24 \)

\( x+5y=4 \)

Question 5

\( x-y=8 \)

\( 2x-2y=16 \)

Exercise #6

−x+y=14 -x+y=14

5x+2y=7 5x+2y=7

Video Solution

Answer

x=−3,y=11 x=-3,y=11

Exercise #7

4x+3y=−11 4x+3y=-11

3x−2y=−4 3x-2y=-4

Video Solution

Answer

x=−2,y=−1 x=-2,y=-1

Exercise #8

4x−8y=16 4x-8y=16

−x−2y=24 -x-2y=24

Video Solution

Answer

x=−10,y=−7 x=-10,y=-7

Exercise #9

6x−2y=24 6x-2y=24

x+5y=4 x+5y=4

Video Solution

Answer

x=4,y=0 x=4,y=0

Exercise #10

x−y=8 x-y=8

2x−2y=16 2x-2y=16

Video Solution

Answer

Infinite solutions

Question 1

\( x-y=8 \)

\( 3x+2y=24 \)

Question 2

Solve the following system of equations:

\( \begin{cases} 5x-y=0 \\ 10x-2y=0 \end{cases} \)

Question 3

\( 5y+3x=15 \)

\( -2y-4x=-34 \)

Question 4

\( 2x-2y=10 \)

\( 4x-4y=32 \)

Question 5

\( x+y=0 \)

\( x+y=10 \)

Exercise #11

x−y=8 x-y=8

3x+2y=24 3x+2y=24

Video Solution

Answer

x=8,y=0 x=8,y=0

Exercise #12

Solve the following system of equations:

{5x−y=010x−2y=0 \begin{cases} 5x-y=0 \\ 10x-2y=0 \end{cases}

Video Solution

Answer

There are infinite solutions.

Exercise #13

5y+3x=15 5y+3x=15

−2y−4x=−34 -2y-4x=-34

Video Solution

Answer

x=10,y=−3 x=10,y=-3

Exercise #14

2x−2y=10 2x-2y=10

4x−4y=32 4x-4y=32

Video Solution

Answer

No solution

Exercise #15

x+y=0 x+y=0

x+y=10 x+y=10

Video Solution

Answer

No solution

Topics learned in later sections

  1. Two linear equations with two unknowns
  2. Algebraic solution for linear equations with two unknowns
  3. Substitution method for two linear equations with two unknowns
  4. Solving with the method of equalization for systems of two linear equations with two unknowns