The Elimination Method

The Elimination Method exercise

• Determine which solution is true for a system of equations.

  Hints

  You can solve this system by using the elimination method.

  In this method, you align two equations so that adding them together will eliminate one variable, making it possible to find the value of the other.

  Notice that the equations have $y$ with an addition sign in one and a subtraction sign in the other. When you add the equations, these $y$ terms will cancel each other out, which means you can eliminate $y$ and solve for $x$ directly.

  Substitution can be used to check if solutions are true. For example, if we had the equation $a+b=8$, and $a=3$ and $b=5$, we can substitute the values in for $a$ and $b$ so the equation is $3+5=8$. This is equal, so we know these values are true!

  Solution

  $\begin{array}{rcl} x+y&=&10\\ x-y&=&4 \end{array}$

  To find the solution, we can add these two equations directly to eliminate $y$ and solve for $x$, then substitute the value of $x$ into one of the equations to solve for $y$.

  Let's solve this system:

  Add the two equations:

  $\begin{array}{rcl} x+y&=&10\\ +(x-y&=&4)\\ \hline 2x&=&14\\ x&=&7 \end{array}$

  Now we substitute $x = 7$ into the first equation:

  $\begin{array}{rcl} 7+y&=&10\\ y&=&3 \end{array}$

  The solution to this system of equations is $x = 7$ and $y = 3$, or $(7,3)$.

• Describe the elimination method for solving systems of linear equations.

  Hints

  The solution to a system of equations should be given as an ordered pair in $(x,y)$ format.

  When multiplying an equation by a constant, all values on both sides of the equal sign must be multiplied by that value.

  For example, if we wanted to change the $-x$ in this equation ($-x+3y=15$) to a positive $x$, the entire equation would be multiplied by $-1$ using the distributive property.

  $(-1)(-x+3y=15)$

  $x - 3y = -15$

  When checking the solution to a system of equations, the ordered pair must make both equations true when substituted in for the values of $x$ and $y$.

  Solution

  We are going to use the elimination method to solve a system of linear equations. In this method, we need to find a variable we can eliminate. We first want to create an additive inverse pair for one of our two variables.

  Step 1: In this case, since the $x$ values have matching coefficients, we can work to eliminate them. To create an inverse pair of values, multiply the second equation by $\bf{-1}$ to get $x-9y=23$.

  $~$

  Step 2: Next, we will need to add the two equations together to eliminate a variable. The $x$ values will add to zero, so they cancel out, resulting in the equation $\bf{-16y=32}$.

  $~$

  Step 3: Now that we have an equation with a single variable, we can use inverse operations to isolate the variable. To isolate the $y$ variable, divide both sides by $\bf{-16}$ to get $y=-2$.

  $~$

  Step 4: Now that we know the value of $y$, we can solve for $x$. We can do this by substituting $\bf{-2}$ for $y$ in one of the original equations.

  • Regardless of which of the original equations you choose, you will get the same result.

  $~$

  Step 5: Substituting $-2$ for $y$ in the first equation gives us $\bf{-x-7(-2)=9}$. Simplifying this equation results in $-x+14=9$. Solve this equation to get $x=5$.

  • When solving the equation, you will need to make sure you end up with a positive x. One way to do this is to divide both sides of the equation by $-1$.

  $~$

  Step 6: The ball and the receiver will intersect at the point $(5,-2)$.

  • Make sure to write the ordered pair as $(x,y)$ not $(y,x)$.

  $~$

  Step 7: To check this solution, substitute $5$ for $x$ and $-2$ for $y$ in both equations. Substituting into the first equation results in $\bf{-5-7(-2)=9}$. After simplifying we can see that the solution is correct for this equation since $9=9$. Substituting into the second equation results in $\bf{-5+9(-2)=-23}$. After simplifying we can see that the solution is correct for this equation since $-23=-23$.

• Order the steps for solving a system of linear equations through elimination.

  Hints

  The first step requires one of the equations to be manipulated in order to cancel out one of the variables.

  In this case, it is easier to eliminate the $x$ values.

  Solve for $y$ prior to solving for $x$.

  Solution

  • Multiply the first equation by $3$ to get a common multiple for the coefficients of $x$.

  $3(-x+7y=2)$

  $-3x+21y=6$

  • Add the equations:

  $\begin{array}{ccc} -3x+21y&=&6\\ 3x+5y&=&46\\ \text{to get}\\ 26y=52\\ \end{array}$

  • Solve for $y$:

  $y=2$

  • Substitute $2$ in for $y$ in the first equation:

  $-x+7(2)=2$

  • Solve for $x$:

  $x=12$

  • The receiver and the ball will intersect at the point $(12,2)$.

• Apply the steps to solve a systems of equations with elimination method.

  Hints

  One solution: This type of system has lines that intersect at one point, and using elimination will give you a single pair of $x$ and $y$ values that solve both equations.

  No solution: The lines are parallel and never meet; elimination will lead to a false statement, like $0 = 5$, indicating that there is no common solution.

  Infinite solutions: The lines overlap exactly, and elimination will result in a true statement regardless of $x$ and $y$, like $0 = 0$, showing that any point on the line solves the system.

  When multiplying an equation by a constant, make sure to multiply everything on both sides of the equation to get an equivalent equation.

  This is known as the Distributive Property.

  When adding an expression with opposite coefficients such as $-4x$ and $4x$, the $x$ values are eliminated, so $-4x+4x=0$.

  Solution

  This system of equations has no solution.

  $\begin{array}{lcr} x+2y&=&4\\ 3x+6y&=&18 \end{array}$

  • Multiply the first equation by $-3$:

  $\begin{array}{lcr} -3(x+2y&=&4)\\ 3x+6y&=&18 \end{array}$

  $~$

  $\begin{array}{lcr} -3x-6y&=&-12\\ 3x+6y&=&18 \end{array}$

  • Add the equations together to notice that the $-3x$ and $3x$, as well as the $-6y$ and $6y$ will cancel each other out, resulting in $0$ on that side of the equal sign.

  $0=6$

  • Because $0$ is not equal to $6$, there is no solution.

• Review evaluating expressions and finding the least common multiple of two numbers.

  Hints

  The least common multiple of two numbers is the smallest number that they both divide evenly into. For example, the least common multiple of $6$ and $8$ is $24$. To get this answer, start by listing some of the multiples of each number.

  $\bf{6}$: $6, 12, 18, 24, 30$

  $\bf{8}$: $8, 16, 24$

  As soon as you find a common multiple, you can stop making your list since we are looking for the least common multiple.

  Use the distributive property to simplify expressions in the form $a(b+c)$. For example, if you have the expression $3(x+4)$, multiply each value inside the parentheses by $3$. This results in $3(x)+3(4)$. This expression simplifies to $3x+12$.

  In order to evaluate an expression, substitute (replace) the given value of the variable into the expression and simplify. For example, to evaluate the expression $x+5$ given that $x=2$, replace the $x$ with $2$ to get $2$$+5 = 7$.

  Solution

  • The least common multiple of $3$ and $7$ is $21$ since $3(7)=21$ and $7(3)=21$

  $~$

  • Evaluating the expression $-3x+4$ for $x=-8$: Substitute $x$ for $-8$ in the expression: $-3(-8)+4 = 24+4 = 28$

  $~$

  • The least common multiple of $3$ and $6$ is $6$ since $3(2)=6$ and $6(1)=6$

  $~$

  • Evaluating the expression $-4x+5$ for $x=\left( \frac{1}{2} \right)$: $-4\left( \frac{1}{2} \right)+5=-2+5=3$

• Solve the real-word problems using systems of linear equations.

  Hints

  Pay careful attention to keywords such as

  • combined total: adding
  • twice: multiplied by $2$
  • more than: adding

  When you are solving using the elimination method, it is helpful to have the variables line up, like this:

  $\begin{array}{rcl} x+3y&=&18\\ -x-4y&=&-25 \end{array}$

  To find the perimeter of a rectangle, add all the sides together. Since there are two pair of equal sides on a rectangle, you will need to add twice the length and twice the width, so formula looks like this:

  $P = 2L+2W$

  Solution

  Last season, two running backs on the college football team rushed for a combined total of $1,050$ yards. Jamal rushed $2$ times as many yards as Titon. How many yards were rushed by each player?

  Let $J =$ the number of yards rushed by Jamal

  Let $T =$ the number of yards rushed by Titon

  The first equation is:

  $J+T=1,050$

  Since Jamal rushed $2$ times as many yards as Titon, the second equation is:

  $J=2T$

  First, we need the two equations to be in the same format, so subtract $2T$ from both sides of the equation to get:

  $J-2T=0$

  In order to eliminate the $J's$, multiply the second equation by $-1$:

  $-1(J-2T=0)$

  $-J +2T=0$

  Add the equations together and solve!

  $3T=1,050$

  $T=350$

  Substituting $T=350$ into the first equation $J+T=1,050$, and solve for $J$:

  $J=700$

  Jamal rushed $700$ yards and Titon rushed $350$ yards.

  $~$

  On Monday, Samuel bought $10$ bottles of coconut water and $5$ mangos for his club meeting for $\$16.50$. It turns out that the mangos were more popular than the coconut water. On Tuesday he bought $5$ bottles of coconut water and $10$ mangos for a total of $\$14.25$. How much was each bottle of coconut water?

  Let $C=$the price of the coconut water

  Let $M=$the price of the mangos

  To get the cost, we need to multiply the number of each item by the price of each item:

  Monday: $10C+5M=16.50$

  Tuesday: $5C+10M=14.25$

  In order to get the $C's$ to cancel, multiply the second equation by $-2$:

  $-2(5C+10M=14.25)$

  $-10C-20M=-28.50$

  Add the equations together and solve for $M$:

  $-15M=-12.00$

  $M=0.8$

  Substituting $M=0.8$ into the original equation for Tuesday and solve for $C$:

  $5C+10(0.8)=14.25$

  $C=1.25$

  Each bottle of coconut water cost $\$1.25$.

  $~$

  The perimeter of a rectangular garden is $62$ feet. The length is $1$ foot more than twice the width. Find the dimensions of the garden.

  We know that the formula for the perimeter of a rectangle is $P=2L+2W$. We also know that $P=62$, so we can substitute 62 for $P$ in the formula:

  $62=2L+2W$

  Since we have two variables in our equation, we will need another equation relating length ($L$) and width ($W$) to solve this problem.

  We are given that The length is $1$ foot more than twice the width so our second equation is:

  $L=1+2W$

  To use the elimination method, we need the two equations to be in the same format. We can subtract $2W$ from both sides on the second equation and flip the first equation to get:

  $L-2W=1$

  $2L+2W=62$

  Add the equations together and solve for the missing variable.

  $3L=63$

  $L=21$

  Using the second equation $L=1+2W$, substitute $21$ for $L$ to solve for $W$:

  $21=1+2W$

  $W=10$

  The length of the rectangle is $21$ feet and the width of the rectangle is $10$ feet.