Wize High School Algebra I Textbook (Common Core) > Systems of Linear Equations

Solving a System of Linear Equations - No Solution or Many Solutions

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No Solution or Many Solutions
So far we have been able to find one (x,y)(x,y)(x,y) solution. But there are two more possibilities.
 
Many Solutions
If both lines are equivalent (the same) then they would completely overlap on a graph.


If you solve by substitution or elimination, at some point both of the variables will disappear and you'll end up with an equation that is always true, such as 5=55=55=5, or 0=00=00=0.

Wize Tip
You'll be able to visably see that the two equations are multiples of one another.

Example
2x−4y+10=01◯x−2y+5=02◯\begin{array}{cc}
2x-4y+10=0&\text{\textcircled 1}\\
x-2y+5=0&\text{\textcircled 2}
\end{array}2x−4y+10=0x−2y+5=0​1◯2◯​

Since 1◯=2×2◯\text{\textcircled 1}=2\times\text{\textcircled 2}1◯=2×2◯, these equations are multiples of one another. So the lines are the same and there are many solutions.

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No solution
If the lines are parallel, they will never meet on the graph.


If you solve by substitution or elimination, at some point both of the variables will disappear and you'll end up with an equation that is always false, such as 5=35=35=3, or −4=0-4=0−4=0.

Wize Tip
You'll be able to visably see that the corresponding variables are the same multiples of one another, but the constant is not.

Example
2x−4y+9=01◯x−2y+5=02◯\begin{array}{cc}
2x-4y+9=0&\text{\textcircled 1}\\
x-2y+5=0&\text{\textcircled 2}
\end{array}2x−4y+9=0x−2y+5=0​1◯2◯​

Notice that the variables in 1◯\text{\textcircled 1}1◯ are 2 times the variables in 2◯\text{\textcircled 2}2◯, but the constant in 1◯\text{\textcircled 1}1◯ is NOT 2 times the constant in 2◯\text{\textcircled 2}2◯. So the lines are parallel and do not mett on the graph, and there are no solutions to this system of linear equations.

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Example: Many Solutions
Solve
10x−5y=−2010x-5y=-2010x−5y=−20
−2x+y=4-2x+y=4−2x+y=4

Graphing

The two lines overlap providing many solutions as every point is a (x,y)(x,y)(x,y) for both lines.

  

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Substitution
10x−5y=−201◯−2x+y=42◯\begin{array}{cc}
10x-5y=-20&\text{\textcircled 1}\\
-2x+y=4&\text{\textcircled 2}
\end{array}10x−5y=−20−2x+y=4​1◯2◯​

From 2◯\text{\textcircled 2}2◯:
−2x+y=4y=2x+4\begin{array}{rcl}
-2x+y&=&4\\

y&=&2x+4
\end{array}−2x+yy​==​42x+4​ 
Substitute this value of y=2x+4\colorbox{yellow}{$y$}=\colorbox{yellow}{$2x+4$}y​=2x+4​ into 1◯\text{\textcircled 1}1◯:

10x−5y=−2010x−5(2x+4)=−2010x−10x−20=−20−20=−20\begin{array}{rcl}
10x-5\colorbox{yellow}{$y$}&=&-20\\\\
10x-5(\colorbox{yellow}{$2x+4$})&=&-20\\\\
10x-10x-20&=&-20\\\\
-20&=&-20
\end{array}10x−5y​10x−5(2x+4​)10x−10x−20−20​====​−20−20−20−20​

Both variables have "disappeared" and we end up with a statement that is always true.

Therefore, the lines are the same and overlap in the graph.

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Elimination
10x−5y=−201◯−2x+y=42◯\begin{array}{cc}
10x-5y=-20&\text{\textcircled 1}\\
-2x+y=4&\text{\textcircled 2}
\end{array}10x−5y=−20−2x+y=4​1◯2◯​

Multiply the second equation by 555, leave the first equation alone:

10x−5y=−2010x-5y=-2010x−5y=−20
5×(−2x+y=4)\colorFour5\times(-2x+y=4)5×(−2x+y=4)

Simplify:
10x−5y=−2010x-5y=-2010x−5y=−20
−10x+5y=20-10x+5y=20−10x+5y=20

Adding the two equations, we get

0=00=00=0

Both variables have "disappeared" and we end up with a statement that is always true.

Therefore, the lines are the same and overlap in the graph.

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Example: No Solutions
Solve
2x−y=−61◯−2x+y=42◯\begin{array}{cc}
2x-y=-6&\text{\textcircled 1}\\
-2x+y=4&\text{\textcircled 2}\\
\end{array}2x−y=−6−2x+y=4​1◯2◯​

Graphing

The two lines are parallel providing no solutions as they will never meet at (x,y)(x,y)(x,y)

  

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Substitution
2x−y=−61◯−2x+y=42◯\begin{array}{cc}
2x-y=-6&\text{\textcircled 1}\\
-2x+y=4&\text{\textcircled 2}\\
\end{array}2x−y=−6−2x+y=4​1◯2◯​

From 2◯\text{\textcircled 2}2◯:
−2x+y=4y=2x+4\begin{array}{rcl}
-2x+y&=&4\\
y&=&2x+4
\end{array}−2x+yy​==​42x+4​

Substitute this value of y=2x+4\colorbox{yellow}{$y$}=\colorbox{yellow}{$2x+4$}y​=2x+4​ into 1◯\text{\textcircled 1}1◯:

2x−y=−62x−(2x+4)=−62x−2x−4=−6−4=−6\begin{array}{rcl}
2x-\colorbox{yellow}{$y$}&=&-6\\\\
2x-(\colorbox{yellow}{$2x+4$})&=&-6\\\\
2x-2x-4&=&-6\\\\
-4&=&-6
\end{array}2x−y​2x−(2x+4​)2x−2x−4−4​====​−6−6−6−6​

Both variables have "disappeared" and we end up with a statement that is always false.

Therefore, the lines are parallel and do not meet on the graph, so there are no solutions.

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Elimination
2x−y=−61◯−2x+y=42◯\begin{array}{cc}
2x-y=-6&\text{\textcircled 1}\\
-2x+y=4&\text{\textcircled 2}\\
\end{array}2x−y=−6−2x+y=4​1◯2◯​

Adding the two equations, we get

0=−20=-20=−2

Both variables have "disappeared" and we end up with a statement that is always false.

Therefore, the lines are parallel and do not meet on the graph, so there are no solutions.

Practice: No Solution or Many Solutions
State whether each of the following systems have no solution or many solutions.

a)x−2y+3=01◯2x−4y+6=02◯\begin{array}{cc}
x-2y+3=0&\text{\textcircled 1}\\
2x-4y+6=0&\text{\textcircled 2}
\end{array}x−2y+3=02x−4y+6=0​1◯2◯​

b) −12x+8y=41◯3x−2y=42◯\begin{array}{cc}
-12x+8y=4&\text{\textcircled 1}\\
3x-2y=4&\text{\textcircled 2}
\end{array}−12x+8y=43x−2y=4​1◯2◯​

c) 9x−15y=−61◯−3x+5y=22◯\begin{array}{cc}
9x-15y=-6&\text{\textcircled 1}\\
-3x+5y=2&\text{\textcircled 2}
\end{array}9x−15y=−6−3x+5y=2​1◯2◯​

d) 9x−15y=−61◯−3x+5y=−22◯\begin{array}{cc}
9x-15y=-6&\text{\textcircled 1}\\
-3x+5y=-2&\text{\textcircled 2}
\end{array}9x−15y=−6−3x+5y=−2​1◯2◯​

a) Enter many or no

b) Enter many or no

c) Enter many or no

d) Enter many or no

I don't know