Solve Systems of Two Linear Equations
Let’s look at the following system of two linear equations. We can see that there is one solution to the system based on the graph.
Let’s look at the following system of two linear equations. We can see that there is one solution to the system based on the graph.
\(\begin{cases} y=\frac{1}{2}x+5 \\ 3x-4y=-17 \end{cases}\)
Unless we had an extremely accurate graph by hand, it might be hard to determine what the intersection point is graphically (and this would only become harder/impossible with ordered pairs that contained other fractions, irrational numbers, etc.). Two methods of solving a system of equations algebraically are substitution and elimination.
Substitution is most efficient when there is at least one equation that is already solved for one of the variables (\(y=\), \(x=\), etc.). Because one of the equations in this system is \(y=\frac{1}{2}x+5\), substitution is an excellent method. You can still solve a system using substitution if one of the equations is not already solved for a variable, but you would need to solve one of the equations for one of the variables first.
To solve the system using substitution, we will substitute \(\frac{1}{2}x+5\) in for \(y\) into the other equation. We will then have one equation that only contains one variable (\(x\)) and can solve that for \(x\). Finally, we can substitute that value into one of the original equations to solve for \(y\).
\(\begin{cases} y=\frac{1}{2}x+5 \\ 3x-4y=-17 \end{cases}\)
Substitution is most efficient when there is at least one equation that is already solved for one of the variables (\(y=\), \(x=\), etc.). Because one of the equations in this system is \(y=\frac{1}{2}x+5\), substitution is an excellent method. You can still solve a system using substitution if one of the equations is not already solved for a variable, but you would need to solve one of the equations for one of the variables first.
To solve the system using substitution, we will substitute \(\frac{1}{2}x+5\) in for \(y\) into the other equation. We will then have one equation that only contains one variable (\(x\)) and can solve that for \(x\). Finally, we can substitute that value into one of the original equations to solve for \(y\).
\(\begin{cases} y=\frac{1}{2}x+5 \\ 3x-4y=-17 \end{cases}\)
\(\begin{align}3x-4(\frac{1}{2}x+5)&=-17 \ & \ &\text{1) Substitute}\ \frac{1}{2}x+5\ \text{in for y into the second equation.}\\3x-2x-20&=-17\ & \ &\text{2) Solve for x.}\\x-20&=-17\\x&=3\\3(3)-4y&=-17 \ & \ &\text{3) Substitute 3 in for x into one of the original equations.}\\9-4y&=-17 \ & \ &\text{4) Solve for y.}\\-4y&=-26\\y&=\frac{13}{2}\end{align}\)
The solution to this system of equations is \(\left(3,\frac{13}{2}\right)\). We can check our solution by substituting the values in for \(x\) and \(y\) in each of the original equations and making sure that this ordered pair is a solution to each of those equations.
Elimination (or linear combination) is another method of solving systems of equations. This is when we add the two equations together so that one of the variables is eliminated. We can then solve that equation for the remaining variable and then substitute that value into one of the original equations to solve for the second variable. In order to use elimination, the two equations must be set up the same way in terms of the location of variables and constants and the coefficients on one of the variables should be opposites of each other (so that this variable is eliminated when the equations are added together). Often, we will need to manipulate one or both of the equations first before using elimination.
To solve this same system with elimination, let’s first rewrite \(y=\frac{1}{2}x+5\) so that it is set up the same way as \(3x-4y=-17\).
\(\begin{align}y&=\frac{1}{2}x+5\\-\frac{1}{2}x+y&=5 & &\text{(Subtract}\ \frac{1}{2}x\ \text{from both sides.)}\\x-2y&=-10 & &\text{(Though not necessary, multiplying by -2 eliminates the fraction and makes the first coefficient positive.)}\end{align}\)
We now have the following system.
\(\begin{cases} x-2y=-10 \\ 3x-4y=-17 \end{cases}\)
If we were to add these equations together right now, neither variable would be eliminated. However, if we multiply the first equation by \(-2\), then the first equation will have a \(4y\) and the second equation will have a \(-4y\) and when added together, the \(y\)’s will be eliminated.
\(\begin{cases} -2(x-2y=-10) \\ 3x-4y=-17 \end{cases}\) (Multiply the first equation by \(-2\).)
\(\begin{cases} -2x+4y=20 \\ 3x-4y=-17 \end{cases}\)
\(x=3\) (Add the two equations together so that one of the variables, \(y\), is eliminated.)
Although this solved for \(x\) immediately, you often will have an equation that you need to solve for \(x\).
Now that we know \(x=3\), we can substitute that in for \(x\) into one of the original equations to solve for \(y\) the same way we did with substitution.
\(3(3)-4y=-17\) (Substitute \(3\) in for \(x\) into one of the original equations.)
\(9-4y=-17\)
\(-4y=-26\)
\(y=\frac{13}{2}\)
The solution to this system of equations is \(\left(3,\large\frac{13}{2}\right)\).
Example 1: Solve the system of equations: \(\begin{cases} 5x-3y=16 \\ -2x+4y=-26 \end{cases}\)
The solution to this system of equations is \(\left(3,\frac{13}{2}\right)\). We can check our solution by substituting the values in for \(x\) and \(y\) in each of the original equations and making sure that this ordered pair is a solution to each of those equations.
Elimination (or linear combination) is another method of solving systems of equations. This is when we add the two equations together so that one of the variables is eliminated. We can then solve that equation for the remaining variable and then substitute that value into one of the original equations to solve for the second variable. In order to use elimination, the two equations must be set up the same way in terms of the location of variables and constants and the coefficients on one of the variables should be opposites of each other (so that this variable is eliminated when the equations are added together). Often, we will need to manipulate one or both of the equations first before using elimination.
To solve this same system with elimination, let’s first rewrite \(y=\frac{1}{2}x+5\) so that it is set up the same way as \(3x-4y=-17\).
\(\begin{align}y&=\frac{1}{2}x+5\\-\frac{1}{2}x+y&=5 & &\text{(Subtract}\ \frac{1}{2}x\ \text{from both sides.)}\\x-2y&=-10 & &\text{(Though not necessary, multiplying by -2 eliminates the fraction and makes the first coefficient positive.)}\end{align}\)
We now have the following system.
\(\begin{cases} x-2y=-10 \\ 3x-4y=-17 \end{cases}\)
If we were to add these equations together right now, neither variable would be eliminated. However, if we multiply the first equation by \(-2\), then the first equation will have a \(4y\) and the second equation will have a \(-4y\) and when added together, the \(y\)’s will be eliminated.
\(\begin{cases} -2(x-2y=-10) \\ 3x-4y=-17 \end{cases}\) (Multiply the first equation by \(-2\).)
\(\begin{cases} -2x+4y=20 \\ 3x-4y=-17 \end{cases}\)
\(x=3\) (Add the two equations together so that one of the variables, \(y\), is eliminated.)
Although this solved for \(x\) immediately, you often will have an equation that you need to solve for \(x\).
Now that we know \(x=3\), we can substitute that in for \(x\) into one of the original equations to solve for \(y\) the same way we did with substitution.
\(3(3)-4y=-17\) (Substitute \(3\) in for \(x\) into one of the original equations.)
\(9-4y=-17\)
\(-4y=-26\)
\(y=\frac{13}{2}\)
The solution to this system of equations is \(\left(3,\large\frac{13}{2}\right)\).
Example 1: Solve the system of equations: \(\begin{cases} 5x-3y=16 \\ -2x+4y=-26 \end{cases}\)
Use the sliders to manipulate the second equation and find values for \(a\) and \(b\) that cause the system to have no solution and infinitely many solutions.
You should have noticed that when the lines were parallel (\(a=-2\) and \(b\) was anything except \(-3\)), then there were no solutions to the system. When the two lines were the same (\(a=-2\) and \(b=-3\)), then there are an infinite number of solutions. If \(a\) is anything except \(-2\) (not parallel), then there will only be one solution.
Let’s look at what happens when we solve these types of systems algebraically.
No Solution
\(\begin{cases} 2x+6y=-15 \\ 3x+9y=4 \end{cases}\)
Because both equations are set up the same way, elimination will be the most efficient method to solve it.
We can multiply the first equation by \(3\) and the second equation by \(-2\) to create equations that when added together will have the \(x\)’s eliminated. However, unlike the other examples we looked at, the \(y\)’s will also be eliminated.
\(\begin{cases} 3(2x+6y=-15) \\ -2(3x+9y=4) \end{cases}\)
\(\begin{cases} 6x+18y=-45 \\ -6x-18y=-8 \end{cases}\)
\(0=-41\) (add the two equations together...both variables are eliminated)
This is a false statement which means that there is NO SOLUTION to this system. Anytime the variables are all eliminated in a system of linear equations and you end with a false statement, there is no solution to the system.
Infinitely Many Solutions
\(\begin{cases} 10x+8y=24 \\ y=-\frac{5}{4}x+3 \end{cases}\)
Because one of the equations is already solved for \(y\), substitution will be the most efficient method to solve it.
\(10x+8(-\frac{5}{4}x+3)=24\) (substitute \(-\frac{5}{4}x+3\) in for \(y\) into the first equation)
\(10x-10x+24=24\) (distribute \(8\))
\(24=24\)
This is a true statement which means that there are INFINITELY MANY SOLUTIONS to this system. Anytime, the variables are all eliminated in a system of linear equations and you end with a true statement, there are infinitely many solutions to the system. Again, note that we do not say all solutions as only the ordered pairs on these lines are solutions.
Quick Check
Solve each system of equations algebraically.
1) \(\begin{cases} y=\frac{1}{3}x+\frac{5}{6} \\ 4x+12y=10 \end{cases}\)
2) \(\begin{cases} 4x+7y=-2 \\ -x-5y=7 \end{cases}\)
3) \(\begin{cases} 6x-4y=7 \\ 3x-2y=-5 \end{cases}\)
Quick Check Solutions