Equation Lab #10 – Systems of Equations: The Substitution Method (1)

During the last two days, I went over the elimination method. On Sunday, I went over the basics, while on Monday, I went over how to set up equations. For the next few days, I would like to cover the other analytic method on solving systems of equations, the substitution method. Instead of combining two equations to eliminate a variable, you’ll have to replace a variable with an algebraic expression.

The Substitution Method – Part 1:

Before I get started with today’s examples, let’s recall variable evaluation. When we know what the value of the variable is, we can replace the variable with the constant. Say like $y+2$. If we found out that $y=5$, then $y+2$ can be rewritten as $5+2$, which simplifies into $7$. If we found out that $y=14$, then $y+2$ can be rewritten as $14+2$, which is also $16$. But what if $y=x$? The expression becomes $x+2$. What if $y=2x+2$? The expression becomes $2x+2+2$, which simplifies into $2x+4$. You can substitute single variables with algebraic expressions.

Here’s another example. If $y=2x+4$ and $y=3x-5$, does that mean $2x+4=3x-5$? The answer is yes. If both $2x+4$ and $3x+5$ equals $y$, then both expressions have the same value. This is how the substitution method works.

Let’s start with our first example: $v=4s-69$, and $v=s+6$. Since $v$ is equal to both $4s-69$ and $s+6$ at the same time, we can write $4s-69=s+6$. We’ll eventually solve for $v$, but first, let’s solve for $s$. If you remember how to solve equations with variables on both sides, you would know that you can subtract $s$ from both sides to isolate the constant on the right. The result is $3s-69=6$. Adding $69$ to both sides, you get $3s=75$. Divide both sides by $3$, you get $s=25$. With $s$ being known, we can find $v$. Let’s plug in $s=25$ in $v=s+6$. You will get $v=25+6$, which in return becomes $v=31$. So when $v=4s-69$ and $v=s+6$, we get that $s=25$ and $v=31$.

 1. $v=4s-69$
    $v=s+6$
 2. $(4s-69)=s+6$
 3. $4s-69=s+6$
 4. $3s-69=6$
 5. $3s=75$
 6. $s=25$
 7. $v=(25)+6$
 8. $v=25+6$
 9. $v=31$
10. $s=25, v=31$

Let’s check our work to see if both $s=25$ and $v=31$ are correct. We know that they fit the second equation since we evaluated the expression rather than solve it. As for the first equation, we get that $31=4(25)-69$. $4\cdot 25=100$, and $100-69=31$. This makes the first equation hold. Since both equations are true, $s=25$ and $v=31$ are the correct solutions.

Let’s do another example. Our two equations are $k+5j=651$ and $k=j+51$. What’s unusual about this system? It seems that the first equation is written in standard form, while the second equation is written in slope-intercept form. But that doesn’t change the technique. Since we know that $k=j+51$, we can substitute $k$ with $j+51$ in the first equation, $k+5j=651$. It becomes $j+51+5j=651$. Combining like terms, we get $6j+51=651$. Subtracting $51$ from both sides, we get $6j=600$. Divide both sides by $6$, we get $j=100$.

Now that we know that $j=100$, let’s substitute $j$ in $k=j+51$. We get $j=100+51$, which is $151$. So when $k+5j=651$ and $k=j+51$, we get that $k=151$ and $j=100$.

 1. $k+5j=651$
    $k=j+51$
 2. $(j+51)+5j=651$
 3. $j+51+5j=651$
 4. $6j+51=651$
 5. $6j=600$
 6. $j=100$
 7. $k=(100)+51$
 8. $k=100+51$
 9. $k=151$
10. $k=151, j=100$

Let’s check our work to see if both $k=151$ and $j=100$ are correct. We know that they fit the second equation since we evaluated the expression rather than solve it. As for the first equation, we get that $(151)+5(100)=651$. $5\cdot 100=500$, and $500+151=651$. Since both equations are true, $k=151$ and $j=100$ are the correct solutions.

Word Problem #9:

The high temperature at 4:00 PM is three times as high as the low temperature at 4:00 AM. The difference is 60°F. What are the high and low points?

Like always, we define our variables. Let’s define $h$ as the high temperature, and $l$ as the low temperature. One equation is the difference. If the difference between both temperatures is 60 degrees Fahrenheit, then the equation is $h-l=60$. This sounds like both samples were taken in the desert. We are also given that the high temperature is three times as high as the low temperature. So the second equation is $h=3l$. Let’s solve.

With $h$ already bare, you can replace $h$ with $3l$ in the equation $h-l=60$. The equation becomes $3l-l=60$. Combining like terms, you get $2l=60$. Divide both sides by $2$, you get $l=30$.

So with $l$ found, let’s plug it in. We’ll choose the equation $h=3l$. All you get to do is to multiply $30$ by $3$. The result is $90$.

 1. Define $h$ as the high temperature,
    and $l$ as the low temperature.
 2. The high temperature is 3 times
    as high as the low temperature,
    and it's 60 degrees higher.
 3. $h-l=60$
    $h=3l$
 4. $(3l)-l=60$
 5. $3l-l=60$
 6. $2l=60$
 7. $l=30$
 8. $h=3(30)$
 9. $h=90$
10. $h=90, l=30$

The high temperature is 90°F. The low temperature is 30°F.

Now if we check our work, $90-60=30$. So either way, you’ll get that $h=90$ and $l=30$. Our answer is correct.

Quiz #10:

Now that the lesson is over, let’s see if you can complete this quiz. Like always, there are eight questions, but there are four systems of equations to solve for. Whether you put down the missing value or the variable with the missing value is fine for as long as you use the correct variable and correct answer. For instance, when the question is $b+3=6$, you can use $3$ or $b=3$ as your answer, but not $a=3$.