Solving Systems of Equations by Substitution

Solving Systems of Equations by Substitution exercise

• Explain how to solve a system of equations by substitution.

  Hints

  Here is an example of a system of equations:

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

  You can substitute $x=y$ into the first equation to get $y+y=4$. This is equivalent to $2y=4$. Dividing by $2$ leads to $y=2$.

  Solution

  Follow these steps for any word problem with more than one variable:

  • First, establish a system of equations by assigning variables to the unknown values and write the given information into equations. Remember: You can solve systems of equations with two variables - as long as you have at least two different equations.
  • Next, we are looking for a variable that is already isolated on one side. We plug in the other side of that equation for the variable in the second equation.
  • Now, we only have one equation with one variable left, which we can solve by isolating the variable.
  • But there is still another variable! To solve for the second variable, we plug in the calculated value for the first variable into the second equation.
  • You can check your solutions by substituting them into both equations.

• Determine the number of long-haired and short-haired cats by substitution.

  Hints

  Follow these steps to solve this system of equations:

  1. First, establish the system of equations.
  2. Look for an equation with an isolated variable and substitute the other side of the equation for the variable in the second equation.
  3. Solve the equation that has only one variable left.
  4. Substitute the solution of the first variable into the equation with two variables and solve for the one you still don't know.

  If Miss Anderson has $4$ long-haired cats, and if she has seven times more short-haired cats, she has $7\times4=28$ short haired cats.

  Use the opposite operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  Solution

  Establish the System of Equations

  • Let's use $x$ for the number of short-haired cats and $y$ for the number of long-haired cats.
  • Because we have $\$1944$ to spend on grooming, and the cost to groom one short-haired cat is $\$18$ and the cost for long-haired cats is $\$36$, we get: $18x+36y=1944$
  • We still need another equation. Miss Anderson has seven times more short-haired cats than long-haired cats, $x=7y$.

  Now we have a system of equations.

  $\begin{align*} 18x+36y&=1944\\ x&=7y \end{align*}$

  Substitution

  Substitute $x=7y$ into the equation $18x+36y=1944$.

  Doing this, we get:

  $18(7y)+36y=1944$

  We have just one variable left, $y$.

  Solve the equation to get $y$

  $\begin{array}{rclcl} 18(7y)+36y&=&1944&&|\text{ multiply}\\ 126y+36y&=&1944&&|\text{ combine like terms}\\ 162y&=&1944 \end{array}$

  Finally, we divide by $162$ to get $y=12$.

  Substitute to get the value for $x$

  We substitute $y=12$ into the equation $x=7y=7\times 12=84$.

  Now we know that Miss Anderson can take $12$ short-haired and $84$ long-haired cats to the cat stylist.

• Decide how many dogs Miss Lovingdogs can bring to the dog stylist.

  Hints

  You have to multiply the cost per dog by the number of dogs. The budget is the sum of the number of poodles and the number of dachshunds.

  We can write a system of equations using the given information:

  • $20x+35y=475$
  • $x=3y$

  Solution

  Establish the system of equations

  • We'll use $y$ to represent the number of poodles and $x$ for the number of dachshunds.
  • Because the cost to style one dachshund is $\$20$, and the cost to style poodles is $\$35$, we get the expression $20x+35y$.
  • Since Miss Lovingdogs has a budget of $\$475$, our final equation is $20x+35y=475$
  • Let's have a look at the other equation: Miss Lovingdogs has three times more dachshunds than poodles, which gives us $x=3y$.

  So now we have enough to solve the system of equations:

  $\begin{align*} 20x+35y&=475\\ x&=3y \end{align*}$

  Substitution

  Substitute $x=3y$ into the first equation $20x+35y=475$:

  $20(3y)+35y=475$

  Now we have one equation with only one variable, $y$.

  Solve the equation to get $y$

  $\begin{array}{rclcl} 20(3y)+35y&=&475&&|&\text{ Multiply}\\ 60y+35y&=&475&&|&\text{ Combine Like Terms}\\ 95y&=&475\\ \color{#669900}{\div95}&&\color{#669900}{\div95}\\ y&=&5 \end{array}$

  Substitute this value for y to get $x$

  We substitute $y=5$ into the second equation:

  $x=3y=3\times 5=15$

  We know now that Miss Lovingdogs can take $5$ poodles and $15$ dachshunds to the dog stylist.

• Write a system of equations for each situation and solve them.

  Hints

  Keep in mind the difference between

  • four times as many as $=\times 4$
  • four more $=+4$.

  Use the Distributive Property if necessary:

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

  You can check each pair of solutions.

  Solution

  Here we have different examples of similar word problems. To set up each of our equations, we multiply the costs of our choices (cookies or lollipops) by the corresponding variables and add the products to equal to the given budget.

  For each of the problems:

  • $x=$ the unknown number of cookies
  • $y=$ the unknown number of lollipops

  (A)

  • Cookie: $\$2$
  • Lollipop: $\$0.50$
  • One more lollipop than cookies.
  • First equation: $2x+0.5y=30$
  • Second equation: $2x=y$

  $\begin{array}{rclcl} 2x+0.5y&=&30\\ 2x+0.5(2x)&=&30&&|\text{ Substitute } 2x \text{ for } y\\ 2x+x&=&30&&|\text{ Distributive Property}\\ 3x&=&30&&|\text{ Combine Like Terms}\\ \color{#669900}{\div~3}&&\color{#669900}{\div~3}&&|\text{ Opposite Operations}\\ x&=&10\\ \\ y&=&2x\\ y&=&2(10)&&|\text{ Plug in } 10 \text{ for } x\\ y&=&20 \end{array}$

  They can buy $10$ cookies and $20$ lollipops.

  $~$

  (B)

  • Cookie: $\$1.50$
  • Lollipop: $\$2$
  • One more lollipop than cookies.
  • First equation: $1.5x+2y=30$
  • Second equation: $x+1=y$

  $\begin{array}{rclcl} 1.5x+2y&=&30\\ 1.5x+2(x+1)&=&30&&|\text{ Substitute } x+1 \text{ for } y\\ 1.5x+2x+2&=&30&&|\text{ Distributive Property}\\ 3.5x+2&=&30&&|\text{ Combine Like Terms}\\ \color{#669900}{-2}&&\color{#669900}{-2}&&|\text{ Opposite Operations}\\ 3.5x&=&28\\ \color{#669900}{\div~3.5}&&\color{#669900}{\div~3.5}&&|\text{ Opposite Operations}\\ x&=&8\\ \\ y&=&x+1\\ y&=&8+1&&|\text{ Plug in } 8 \text{ for } x\\ y&=&9 \end{array}$

  They can buy $8$ cookies and $9$ lollipops.

  $~$

  (C)

  • Cookie: $\$2$
  • Lollipop: $\$0.50$
  • Same number of cookies as lollipops.
  • First equation: $2x+0.5y=30$
  • Second equation: $x=y$

  $\begin{array}{rclcl} 2x+0.5y&=&30\\ 2x+0.5(x)&=&30&&|\text{ Substitute } x \text{ for } y\\ 2x+0.5x&=&30&&|\text{ Distributive Property}\\ 2.5x&=&30&&|\text{ Combine Like Terms}\\ \color{#669900}{\div~2.5}&&\color{#669900}{\div~2.5}&&|\text{ Opposite Operations}\\ x&=&12\\ \\ y&=&x\\ y&=&12&&|\text{ Plug in } 12 \text{ for } x\\ y&=&12 \end{array}$

  They can buy $12$ cookies and $12$ lollipops.

  $~$

  (D)

  • Cookie: $\$1.50$
  • Lollipop: $\$2$
  • Three times as many lollipops as cookies.
  • First equation: $1.5x+2y=30$ (same as (B))
  • Second equation: $3x=y$

  $\begin{array}{rclcl} 1.5x+2y&=&30\\ 1.5x+2(3x)&=&30&&|\text{ Substitute } 3x \text{ for } y\\ 1.5x+6x&=&30&&|\text{ Distributive Property}\\ 7.5x&=&30&&|\text{ Combine Like Terms}\\ \color{#669900}{\div~7.5}&&\color{#669900}{\div~7.5}&&|\text{ Opposite Operations}\\ x&=&4\\ \\ y&=&3x\\ y&=&3(4)&&|\text{ Plug in } 4 \text{ for } x\\ y&=&12 \end{array}$

  They can buy $4$ cookies and $12$ lollipops.

• Determine the two equations that are needed to correctly describe Miss Anderson's problem.

  Hints

  If the number of long haired cats is $3$, the number or short haired cats is $7\times 3=21$. Which equation reflects this information?

  Solution

  First, we write a system of equations using the given information.

  We'll use $x$ for the number of short-haired cats and $y$ for the number of long-haired cats.

  Next, we take a look at the costs:

  • Miss Anderson has a budget of $\$1944$.
  • The styling cost for short-haired cats is $\$18$ and $\$36$ for long-haired cats.
  • The equation to describe this situation is $18x+36y=1944$.

  We also know that Miss Anderson has seven times more short-haired cats than long-haired cats. This information, written mathematically, is $x=7y$.

  So we have the following system of equations:

  $\begin{align*} 18x+36y&=1944\\ x&=7y \end{align*}$

• Solve the system of equations.

  Hints

  Divide the second equation by $2$ to isolate $y$.

  Substitute $2x$ for $y$ in the first equation.

  To solve an equation with only one variable:

  • First combine like terms
  • Then use opposite operations

  Use the opposite operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  Solution

  To be able to substitute one variable into the other equation, we need to have the variable isolated on one side of the equation.

  Let's take a look at the second equation $4x=2y$. Dividing by $2$ gives us $2x=y$.

  Great, this looks familiar. Next, we substitute $2x$ for $y$ in the first equation, $4x+6y=192$:

  $4x+6(2x)=192$

  Simplifying and combining like terms gives us

  $16x=192$

  Finally, we divide by $16$ to get $x=12$.

  We substitute $12$ for $x$ into the modified equation $y=2x=2\times 12=24$.

  To be sure, we check the solution:

  $\begin{align*} 4x+6y&=192\\ 4\times 12+6\times 24&=192\\ 48+144&=192\\ 192&=192~\surd\\ \\ 4x&=2y\\ 4\times 12&=2\times 24\\ 48&=48~\surd \end{align*}$

  Great!