Equation Lab #8 – Systems of Equations: The Elimination Method (1)

As the month is about to come to an end, I would like to share the Equation Lab lessons of the month. Last month, I went over how to solve single-variable linear equations and multi-step linear equations. While this week will be about linear equations again, it’s going to cover systems of linear equations.

Introduction:

Last month, you learned how to solve equations like this:

• $2x+4=10$

Now what about equations with two variables, like $2x+y=10$? In cases like this, you can only solve for one of the variables, and you can’t tell what definite value fits that variable. But when are expected to find the definite value of the variables, you are given two equations. So one of the equations is $2x+y=10$, and the other is $y=x+1$.

This is what we call a system of equations. You start out with solving for one variable by using both equations. Once you find the value of that variable, you can plug it into either equation and solve for the other. Once you have the values for all variables, you have solved the system.

• When there are two variables to solve for, there are two equations.
• When there are three variables to solve for, there are three equations.
• When there are four variables to solve for, there are four equations.

For this week, we’ll focus on systems of two equations and systems of three equations, all of them being linear equations.

There are six ways to solve a linear system:

• Graphing – if you have a graphing calculator or graph paper, you can graph the equations of the lines. If you know how to graph one lines, you can also graph two lines. Although this will be discussed in a later entry, there are three cases of what if you solved by graphing:
  • Different Slopes – if two lines do not have the same slope, they will become intersecting lines. Both lines will only meet at one point. That point of intersection is the solution. In these cases, they have only one point of intersection.
  • Same Slopes, Same y-Intercepts – if two lines have both the same slope and same y-intercept, they will become coincident lines. Both lines will share every point in common. In these cases, they have infinitely many solutions, thus no unique solution.
  • Same Slopes, Different y-Intercepts – if two lines have the same slope, but different y-intercepts, they will become parallel lines. Parallel lines do not have any points in common. In these cases, they have no solution.
• Substitution – one of the analytic ways of solving systems, you can replace a variable with an algebraic expression the variable is equal to in the other equation, resulting in only one variable in the equation.
• Elimination – another analytic way of solving systems, you can combine two equations, but eliminate one of the variables in the process, giving you only one variable to solve with.
• Inverse Matrix Method – you set up a square matrix that represents the coefficients, and a rectangular matrix that represents the constants on the right side. You find the inverse of the square matrix, then multiply it to the rectangular matrix, and then get your solution.
• Cramer’s Rule – you set up an augmented matrix where the square matrix on the left is the coefficients, and the rectangular matrix on the right is the constants. The x-matrix substitutes the x-coefficients with the constants, and the y-matrix substitutes the y-coefficients with the constants. You then find the determinants of the original matrix and the modified matrices. Then you divide each modified matrix determinant by the original matrix determinant to get your values for each variable.
• Gaussian Elimination – you set up an augmented matrix where the square matrix on the left is the coefficients, and the rectangular matrix on the right is the constants. You apply row operations by addition, subtraction, and multiplication, until you have 1’s in every pivot entry (entry 11, entry 22, entry 33 etc.).
  • In Gaussian Form, all entries in front of the pivots have a value of $0$, while the pivot entries are $1$.
  • In Gauss-Jordan Form, all entries that are not pivot entries have a value of $0$, while the pivot entries are $1$.

For this week, we’re only going to cover substitution and elimination, as well as a spin-off of Cramer’s Rule.

The Elimination Method – Part 1:

The first two days this week will cover the elimination method, where you combine two equations and toss out the variable.

Here is a diagram of how to combine two equations:

• System of Equations:
  • $A_{1}x+B_{1}y=C_{1}$
  • $A_{2}x+B_{2}y=C_{2}$
• Combined Equation:
  • $(A_{1}+A_{2})x+(B_{1}+B_{2})y=C_{1}+C_{2}$

Basically, if you add two equations, you must combine like terms, as the combo equation is the result of adding two equations together. You can also subtract equations.

Now if two of the coefficients to the same variable are additive inverses, combining the two equations will eliminate the variable, giving you only one variable to work with. Likewise, if they are the same coefficients, subtracting the two equations will eliminate the variable.

Let’s start with these two equations: $4c+3d=38$ and $5c-3d=7$. We’ll start with solving for $c$, which means you’ll have to eliminate $d$. It looks like both $3d$ and $-3d$ are opposites. So if we combine both equations, $3d$ and $-3d$ become $0$, knocking $d$ out of the equation. But $4c+5c=9c$, and $38+7=45$. So our new equation is $9c=45$.

Using math skills from the previous lessons, this is a multiplicative equation, which can be solved by division. Since the coefficient of $c$ is $9$, we’ll have to divide both sides by $9$. $45/9=5$, which means that $c=5$.

Now that we know what $c$ is, let’s find $d$. You can choose either $4c+3d=38$ or $5c-3d=7$ to plug in $c=5$. We’ll go with the first equation. $4c+3d=38$ is the same as $4(5)+3d=38$, which becomes $20+3d=38$. Subtract $20$ from both sides, you get $3d=18$. Divide both sides by $3$, you get $d=6$. So when $4c+3d=38$ and $5c-3d=7$, we get that $c=5$ and $d=6$.

1. $4c+3d=38$
   $5c-3d=7$
2. $(4c+5c)+(3d-3d)=38+7$
3. $9c=45$
4. $c=5$
5. $4(5)+3d=38$
6. $20+3d=38$
7. $3d=18$
8. $d=6$
9. $c=5$, $d=6$

Let’s check our work to see if both $c=5$ and $d=6$ are correct. In the first equation, you should have $4(5)+3(6)=38$. $4\cdot 5=20$, and $3\cdot 6=18$. You will then have $20+18$, which equals $38$. So the first equation holds. In the second equation, you should have $5(5)-3(6)=7$. We know that $3\cdot 6=18$, but $5\cdot5=25$. $25-18=7$, which is the value on the right. So the second equation holds. Since both equations are true, $c=5$ and $d=6$ are the correct solutions.

Here is the next system of equations: $b+4g=23$, and $b+g=11$. We’ll start with solving for $g$, which means you’ll have to eliminate $b$. Since all four coefficients are positive, we’ll have to subtract the equations. $b-b=0$, so $b$ is all gone. But $4g-g=3g$, and $23-11=12$. The equation is $3g=12$. All we have left to do is to divide both sides by $3$. The result is $g=4$.

With $g$ found, let’s find $b$. To use less multiplication, let’s choose $b+g=11$. Substituting $g$ with $4$, the equation becomes $b+4=11$. Subtracting $4$ from both sides, the result is $b=7$. So when $b+4g=23$ and $b+g=11$, we get that $b=7$ and $g=4$.

1. $b+4g=23$
   $b+g=11$
2. $(b-b)+(4g-g)=23-11$
3. $3g=12$
4. $g=4$
5. $b+(4)=11$
6. $b+4=11$
7. $b=7$
8. $b=7$, $g=4$

Let’s check our work to see if both $b=7$ and $g=4$ are correct. In the first equation, you should have $7+4(4)=23$. $4\cdot 4=16$, but if you add $7$, you get $23$. So the first equation holds. In the second equation, you should have $7+4=11$. The second equation holds since $7+4=11$. Since both equations are true, $b=7$ and $g=4$ are the correct solutions.

With enough coverage, let’s do a word problem.

Word Problem #7:

A fruit bowl has more apples than oranges. The total number of fruits is 18. The difference between apples and oranges is 4. How many apples and oranges are in the bowl?

Let’s define our two variables. We’ll use $a$ for the number of apples, and $o$ for the number of oranges. Now we must write the equations. If the total is 18, but the difference is 4, the first equation should be a sum of the two variables while the second is the difference. This makes the first equation $a+o=18$. The second equation, since there are more apples, is $a-o=4$.

Now we can solve the equations. We’ll first find out how many apples are in the bowl, so we’ll have to eliminate $o$ from the system. Combining $a+o=18$ and $a-o=4$, you should get $2a=22$. Divide both sides by $2$, you will get $a=11$.

Going into the next variable, we can substitute $a$ with $11$ in one of the equations. Let’s choose $a+o=18$. You should get $11+o=18$. Subtracting $11$ from both sides, you will get $o=7$. So when $a=11$, $o=7$.

 1. Define $a$ as the number of apples,
    and define $o$ as the number of
    oranges.
 2. The sum of all fruits is $18$, and
    their difference is $4$.
 3. $a+o=18$
    $a-o=4$
 4. $(a+a)+(o-o)=18+4$
 5. $2a=22$
 6. $a=11$
 7. $(11)+o=18$
 8. $11+o=18$
 9. $o=7$
10. $a=11$, $o=7$

There are 11 apples and 7 oranges in the bowl.

Checking our work, $11+7=18$, but $11-7=4$. Since the sum is still $18$ and the difference is still $4$, our answer is correct.

Quiz #8:

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 $a=3$.