Solving Systems of Equations by Elimination

Solving Systems of Equations by Elimination exercise

• Examine the following statements.

  Hints

  Let's have a look at the following example:

  $\begin{align*} x+y&=4\\ x-y&=2 \end{align*}$

  If we add these equations, we can eliminate the variable y. So we get:

  $2x=6$

  Remember: Usually you need as many equations as you have variables.

  You can multiply any equation with any number (exept for zero).

  Solution

  Because we have the unknown number of containers of sample A as well as the unknown number of containers of sample B, we write equations with two variables.

  Because we have two equations with the same two variables, we have a system of equations:

  $x+y=25$ (this equation represents the total number of containers)

  $3x+7y=115$ (this equation represents the maximum weight)

  For example:

  $\begin{align*} x+y&=4\\ x-y&=2 \end{align*}$

  If we add the equations we get $2x=6$. This gives us one equation and one variable. We can solve for x by using opposite operations. Dividing by $2$ leads to $x=3$. This value can be plugged into $x+y=4$ to solve for y:

  $\begin{array}{rcr} 3+y&=&4\\ -3&=&-3\\ y&=&1 \end{array}$

• Describe how to solve systems of equations by elimination.

  Hints

  Multiply one or both equations so that the coefficients of either $x$ or $y$ are the same except for their signs.

  Use the Distributive Property

  $a\times(b+c)=a\times b+a\times c$.

  After multiplying, if you add one equation to the other, you can eliminate one equation as well as one variable.

  Solve this equation with one variable using opposite operations.

  Then plug the value of the known variable into one of the original equations and solve for the second variable.

  Solution

  Consider the coefficients of $x$ and $y$ in each equation.

  Be strategic when determining how to multiply the equation in order to get common coefficients, for example for $x$, but with different signs. So we multiply the first equation by $-3$:

  $\begin{align*} -3\times(x+y)&=-3\times 25\\ 3x+7y&=115 .\end{align*}$

  Using the Distributive Property this gives us:

  $\begin{align*} -3x-3y&=-75\\ 3x+7y&=115. \end{align*}$

  Well done, the coefficient of $x$ in the first equation is $-3$ and the coefficients of x in the second equation is $3$.

  We add the equations and get:

  $4y=40$

  Next we divide by $4$ to solve for $y$:

  $y=10$

  Plug $y=10$ into the first equation:

  $x+10=25$

  Now, we subtract $10$ on both sides of the equation to get $x=15$.

• Solve the given systems of equations by elimination.

  Hints

  In each of the given systems of equations you can add the equations without having to multiply first.

  After adding the equations, you still have one equation and one variable. You can solve this equation using opposite operations.

  The opposite operation of multiplication is division, and vice versa.

  Don't forget the second variable.

  It doesn't matter into which of the original equations you plug in the solution of the first variable.

  Solution

  So, let's practice elimination:

  1. Add the equations to eliminate one variable.
  2. Solve this equation using opposite operations.
  3. Plug in the solution into any of the original equations to get the solution of the second variable.

  (A)

  $\begin{array}{rcr} 3x-2y&=&7\\ 4x+2y&=&14 \end{array}$

  1. Add the equations $\rightarrow$ $7x=21$.
  2. Divide by $7$ $\rightarrow$ $x=3$.
  3. Substitute $x=3$ into $3x-2y=7$:

  $\begin{array}{rcr} 3\times 3-2y&=&7\\ 9-2y&=&7\\ \color{#669900}{-9}&&\color{#669900}{-9}\\ -2y&=&-2 \end{array}$

  Dividing by $-2$ gives us $y=1$.

  $~$

  (B)

  $\begin{array}{rcr} -4x+7y&=&3\\ 4x+2y&=&6 \end{array}$

  1. Add the equations $\rightarrow$ $9y=9$.
  2. Divide by $9$ $\rightarrow$ $y=1$.
  3. Substitute $y=1$ into $4x+2y=6$:

  $\begin{array}{rcr} 4x+2\times 1&=&6\\ 4x+2&=&6\\ \color{#669900}{-2}&&\color{#669900}{-2}\\ 4x&=&4 \end{array}$

  Dividing by $4$ gives us $y=1$.

  $~$

  (C)

  $\begin{array}{rcl} 12x+4y&=32\\ 3x-4y&=13 \end{array}$

  1. Add the equations $\rightarrow$ $15x=45$.
  2. Divide by $15$ $\rightarrow$ $x=3$.
  3. Substitute $x=3$ into $12x+4y=32$:

  $\begin{array}{rcr} 12\times 3+4y&=&32\\ 36+4y&=&32\\ \color{#669900}{-36}&&\color{#669900}{-36}\\ 4y&=&-4 \end{array}$

  Dividing by $4$ gives us $y=-1$.

  $~$

  (D)

  $\begin{array}{rcl} 3x+11y&=&22\\ -3x-4y&=&13 \end{array}$

  1. Add the equations $\rightarrow$ $7y=35$.
  2. Divide by $7$ $\rightarrow$ $y=5$.
  3. Substitute $y=5$ into $3x+11y=22$:

  $\begin{array}{rcr} 3x+11\times 5&=&22\\ 3x+55&=&22\\ \color{#669900}{-55}&&\color{#669900}{-55}\\ 3x&=&-33 \end{array}$

  Dividing by $3$ gives us $x=-11$.

• Determine how much pocket money Ben and Sam get each day.

  Hints

  First you need to have a system of equations: two equations and two variables.

  1. We add the equations or subtract one equation from the other to eliminate one variable.
  2. We solve this equation using opposite operations.
  3. We plug this solution into any of the original equations to get the solution of the other variable.

  Solution

  First we have to set up a system of equations according to the described situation.

  We know that together each week, Sam and Ben get daily pocket money equal to 56 dollars a week. $x$ represents Sam's daily pocket money, and $y$ represents Ben's daily pocket money for a normal week. This gives us the first equation:

  $7x+7y=56$

  Because Sam gets double his pocket money in the week his granny's visit, Sam and Ben have a total of 91 dollars at the end of this week. This gives us the second equation:

  $7(2x)+7y=91$ or simplified:

  $14x+7y=91$.

  How can we solve this system of equations?

  1. We subtract one of the equations from the other to eliminate one variable.
  2. We solve this equation using opposite operations.
  3. We plug this solution into any equation to get the solution of the second variable.

  So, let's start:

  $\begin{align*} 7x+7y&=56\\ 14x+7y&=91. \end{align*}$

  1. Subtracting the equations $\rightarrow$ $7x=35$.
  2. Divide by $7$ $\rightarrow$ $x=5$.
  3. Substitute $x=5$ into $7x+7y=56$:

  $\begin{array}{rcr} 7\times 5+7y&=&56\\ 35+7y&=&56\\ \color{#669900}{-35}&&\color{#669900}{-35}\\ 7y&=&21 \end{array}$

  Now we divide by $7$ and get the solution:

  $y=3$.

  So, for a normal week, Sam gets a daily pocket money of 5 dollars, and Ben gets a daily pocket money of 3 dollars.

• Establish a system of equations for the given situation.

  Hints

  If you have 4 cats and 5 dogs, you have 9 animals in total.

  Remember: Usually you need as many equations as the number of variables.

  Solution

  First, we have to write a system of equations with the given information.

  We assign $x$ to the number of containers of sample A, and $y$ to the number of containers of sample B.

  The total number of containers is 25. This gives us the first equation:

  $x+y=25$

  Next consider the weight of each container:

  • The maximum weight limit is 115 tons
  • $x$ containers of sample A have a weigh 3 tons each
  • $y$ containers of sample B have a weigh 7 tons each

  This gives us the second equation:

  $3x+7y=115$

  Therefore, the system of equations is:

  $\begin{align*} x+y&=25\\ 3x+7y&=115 \end{align*}$

• Solve the system of equations by elimination.

  Hints

  It doesn't matter, which variable you eliminate. The solution is the same.

  It's possible to eliminate one variable by multiplying only one equation.

  If the coefficients of one variable are the same with different signs, you can add the equations. This way you'll get one equation with one variable.

  You can solve this equation by using opposite operations:

  The opposite operation of multiplication is division, and vice versa.

  Solution

  Let's have a look at the coefficients: $8$ and $4$ for $x$, and $13$ and $-4$ for $y$.

  You can eliminate either $x$ or $y$. Because $8$ is $2\times 4$ it's easier to eliminate $x$. Therefore, we multiply the second equation by $-2$:

  $\begin{align*} 8x+13y&=55\\ -2\times(4x-4y)&=-2\times(-4) \end{align*}$

  Using the Distributive Property we get

  $\begin{align*} 8x+13y&=55\\ -8x+8y&=8 \end{align*}$

  Now we add the equations and get:

  $21y=63$

  Dividing by $21$ gives us the solution for $y$:

  $y=3$

  Now we plug this value into any of the original equations, for example $8x+13y=55$:

  $\begin{array}{rcr} 8x+13\times 3&=&55\\ 8x+39&=&55\\ \color{#669900}{-39}&&\color{#669900}{-39}\\ 8x&=&16 \end{array}$

  There is only one step left: We divide by $8$ and get the solution for $x$:

  $x=2$