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Consistent and Inconsistent Linear Systems

Practice Consistent and Inconsistent Linear Systems
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Solving Systems with No or Infinitely Many Solutions Using Substitution

Paran's cell phone company charges a flat rate of $25 per month plus $0.25 per text. Marcel's cell phone company charges a flat rate of $100 and $1 per text. Marcel's bill for the month is four times Paran's. If they sent the same number of texts, how many did they each send?


When a system has no solution or an infinite number of solutions and we attempt to find a single, unique solution using an algebraic method, such as substitution, the variables will cancel out and we will have an equation consisting of only constants. If the equation is untrue as seen below in Example A, then the system has no solution. If the equation is always true, as seen in Example B, then there are infinitely many solutions.

Example A

Solve the system using substitution:

3x-2y &= 7\\y &= \frac{3}{2}x+5

Solution: Since the second equation is already solved for y , we can use this in the first equation to solve for x :

3x-2 \left(\frac{3}{2}x+5\right) &= 7\\3x-3x+5 &= 7\\5 & \ne 7

Since the substitution above resulted in the elimination of the variable, x , and an untrue equation involving only constants, the system has no solution. The lines are parallel and the system is inconsistent.

Example B

Solve the system using substitution:

-2x+5y &= -2\\4x-10y &= 4

Solution: We can solve for x in the first equation as follows:

-2x &= -5y-2\\x &= \frac{5}{2}y +1

Now, substitute this expression into the second equation and solve for y :

4\left( \frac{5}{2} y+1 \right)-10y &=4\\10y +4 -10y &= 4\\4 &=4\\(0 &= 0)

In the process of solving for y , the variable is cancelled out and we are left with only constants. We can stop at the step where 4 = 4 or continue and subtract 4 on each side to get 0 = 0. Either way, this is a true statement. As a result, we can conclude that this system has an infinite number of solutions. The lines are coincident and the system is consistent and dependent.

Example C

Solve the following system using substitution.

2x-3y&=8 \\6x-9y&=24

Solution: Before we begin, first, notice that the second equation is a multiple of the first. Each term is multiplied by 3. Therefore, we know that they are the same equation and will coincide. This system has infinitely many solutions.

Intro Problem Revisit The system of linear equations represented by this situation is:

25 + 0.25x &= y \\100 + x &= 4y

Using substitution, we get:

100 + x &= 4(25 + 0.25) \\100 + x = 100 + x \\0 = 0

There are an infinite number of solutions, so it can't be determined exactly how many texts Paran and Marcel sent.

Guided Practice

Solve the following systems using substitution. If there is no unique solution, state whether there is no solution or infinitely many solutions.

1. y &= \frac{2}{5}x-3\\2x-5y &=15

2. -x+7y &= 5\\3x-21y &= -5

3. 3x-5y &= 0\\-2x+6y &=0


1. Substitute the first equation into the second and solve for x :

2x-5\left(\frac{2}{5}x-3\right) &= 15\\2x-2x+15 &= 15\\15 &=15\\(0 &= 0)

Since the result is a true equation, the system has infinitely many solutions.

2. Solve the first equation for x to get: x=7y-5 . Now, substitute this into the second equation to solve for y :

3(7y-5)-21y &= -5\\21y-15-21y &= -5\\-15 &\ne -5

Since the result is an untrue equation, the system has no solution.

3. Solving the second equation for x we get: x=3y . Now, we can substitute this into the first equation to solve for y :

3(3y)-5y &=0\\9y -5y &= 0\\4y &= 0\\y &= 0

Now we can use this value of y to find x :

x &= 3y\\x &=3(0)\\x &=0

Therefore, this system has a solution at (0, 0). After solving systems that result in 0 = 0, it is easy to get confused by a result with zeros for the variables. It is perfectly okay for the intersection of two lines to occur at (0, 0).


Solve the following systems using substitution.

  1. .
17x - 3y &= 5\\y &= 3x+1
  1. .
4x - 14y &= 21\\y &= \frac{2}{7}x+7
  1. .
-24x + 9y &= 12\\8x-3y &= -4
  1. .
y &= - \frac{3}{4}x+9\\6x + 8y &= 72
  1. .
2x + 7y &= 12\\y &= - \frac{2}{3}x+4
  1. .
2x &= -6y +11\\y &= - \frac{1}{3}x+7
  1. .
\frac{1}{2} x - \frac{4}{5} y &= 8\\5x -8y &= 50
  1. .
-6x + 16 y &= 38\\x &= \frac{8}{3}y - \frac{19}{3}
  1. .
x &= y\\5x +3y &= 0
  1. .
\frac{1}{2}x +3y &= -15\\y &= x-5
  1. .
\frac{2}{3}x +\frac{1}{6}y &= 2\\y &= -4x+12
  1. .
16x -4y &= 3\\y &= 4x+7
  1. .
x-4y &= 10\\8y-2x&=-30
  1. .
4x+5y &= 3\\12x+15y&=9
  1. .




In algebra, to substitute means to replace a variable or term with a specific value.

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