Systems of Equations – Word Problems

Systems of Equations – Word Problems exercise

• Determine the costs of supplies for cats as well as dogs.

  Hints

  Here is an example for combining like terms:

  $34a+17a=(34+17)a=51a$.

  If one scoop of ice cream costs 1.5 dollars, then three scoops costs $3\times 1.5=4.5$ dollars.

  So $x$ scoops of ice cream costs $1.5x$ dollars.

  An example of solving equations using opposite operations:

  Solution

  Jessica wants to know the cost of supplies for cats and for dogs, given that she only has $640$ dollars to spend.

  First, we assign variables to the unknown values:

  • $c$ for the amount of money she can spend per cat.
  • $d$ for the amount of money she can spend per dog.

  As there are $16$ cats and $24$ dogs in the shelter, the total cost of supplies is then $16c+24d$.

  Taking into account the amount of money Jessica has at her disposal, we can see that the amount of money she can spend per cat $c$ and the amount of money she can spend per dog $d$ must satisfy the equation: $16c+24d$$=640$.

  As supplies for a dog are twice as expensive as for a cat, we have that: $d=2c$.

  Now we can substitute $2c$ for $d$ in the equation corresponding to the total cost to get: $16c+48c=640$. We can combine like terms to get: $64c=640$.

  Lastly, we divide by $64$ to get $c=10$. Plugging this into the equation $d=2c$ gives us $d=20$.

• Solve for the number of cats Jessica could buy supplies for.

  Hints

  The number of cats in which supplies can be purchased for multiplied by $10$ gives the amount of money left over after buying supplies for dogs.

  To get the amount of money left over after purchasing dog supplies, subtract the cost of dog supplies from the amount of money Jessica has to spend; i.e. $640$ dollars.

  Solution

  What do we know?

  • Jessica has $640$ dollars at her disposal.
  • Supplies for one cat costs $10$ dollars.
  • She has to buy supplies for $24$ dogs.
  • The special offer is $210$ dollars for supplies for $12$ dogs.

  Let $d$ be the price of supplies for one dog and $x$ the unknown number of cats in which Jessica can buy supplies for if she takes the special offer for dogs.

  So we can establish the following linear equations:

  • $10x+24d=640$, for the total costs.
  • $12d=210$.

  First, we calculate the cost of supplies for one dog by dividing the second equation by $12$ to get $d=17.5$ dollars.

  Because we know that supplies for $12$ dogs cost $210$ dollars, we can conclude that the price for $24$ dogs is $420$ dollars.

  Subtracting $420$ from $640$, we get that the amount of money left over for cat supplies is $640-420=220$ dollars.

  We already know that the price of supplies for one cat is $10$ dollars. So lastly we divide $220$ by $10$ to get $220\div10=22$, the number of cats Jessica has money to buy supplies for after she buys supplies for 24 dogs for the special offer price.

• Describe how to solve a system of equations by graphing.

  Hints

  Slope-intercept form is given by $y=mx+b$, where $m$ is the slope and $b$ is the $y$-intercept.

  Establish the equation describing the total cost.

  Keep in mind that Jessica wants to buy two cat trees. This gives us $2y$.

  The intersection of the lines corresponding to the equations in the system is the solution.

  Solution

  First, we assign variables to what is unknown:

  • Let $x$ be the price of one dog basket.
  • Let $y$ be the price of one cat tree.

  Next, we establish the equations corresponding to the given information.

  Jessica has $48$ dollars in total. She wants to buy two cat trees and one dog basket. This gives us the equation $2y+x=48$.

  One cat tree $y$ is one and a half times as expensive as one dog basket $x$. So we have $y=1.5x$; this equation is in slope-intercept form.

  Transform the first equation to get it into slope-intercept form as well:

  $\begin{array}{rcl} 2y+x&=&48\\ \color{#669900}{-x} & &\color{#669900}{-x}\\ 2y & = & -x+48\\ \color{#669900}{\div2} & & \color{#669900}{\div2}\\ y & = & -\frac12 x+24 \end{array}$

  You can see the corresponding lines above. The intersection is given by $x=12$ and $y=18$.

  Let's check algebraically that this is indeed the solution:

  • $2\times 18+12=36+12=48$ $~~~~~$✓
  • $1.5\times 12=18$ $~~~~~$✓

• Determine the price of dog as well as cat supplies.

  Hints

  Remember to use substitution.

  By eliminating one variable, you get one equation in one variable; use opposite operations to solve this equation.

  Check your solution: both equations must be satisfied.

  Solution

  Let's start with the first equation:

  We substitute $1.5x$ for $y$ in the equation $12x+8y=240$ to get $12x+8(1.5x)=240$. Combining like terms leads to $24x=240$. Lastly, we divide by $24$ to get $x=10$. Again using $y=1.5x$ we can conclude that $y=15$. Let's check that the solution works:

  • $12\times 10+8\times 15=120+120=240$ $~~~~~$✓
  • $1.5\times 10= 15$ $~~~~~$✓

  In a similar manner we solve the next systems of equations:

  Substitute $2.5x$ for $y$ in the equation $12x+8y=128$ to get $12x+8(2.5y)=128$. Combining like terms gives us $32x=128$. Dividing by $32$ results in $x=4$. Multiplying by $2.5$ leads to $y=10$. Once again we check the solution:

  • $12\times 4+8\times 10=48+80=128$ $~~~~~$✓
  • $2.5\times 4= 10$ $~~~~~$✓

  Substitute $2y$ for $x$ in the equation $5x+5y=180$ to get $5(2y)+5y=180$. Combining like terms we get $15y=180$. Dividing by $15$ gives us $y=12$. Multiplying by $2$ leads to $x=24$. And lastly we check the solution:

  • $5\times 24+5\times 12=120+60=180$ $~~~~~$✓
  • $2\times 12= 24$ $~~~~~$✓

• Explain how to solve systems of equations.

  Hints

  Here is an example of a system of equations:

  • $2x+3y=7$
  • $x=2y$

  In the system of equations given in the first hint, $x$ equals $2y$. So plug in $2y$ for $x$ in the first equation to get $2(2y)+3y=7$.

  Keep in mind that you have to find solutions which satisfy all equations.

  Solution

  A system of equations is not only one equation with one variable, but two equations with two variables, or more equations with more variables!

  Remember that solutions to a system of equations must satisfy every equation in the system. Not just one equation in the system.

  How can we solve a system of equations?

  We can eliminate one variable; let's have a look at the following example:

  • $2x+3y=7$
  • $x=2y$

  Substituting $2y$ for $x$ in the first equation gives $4y+3y=7y=7$. Dividing by $7$ gives us $y=1$, and thus $x=2\times 1=2$.

  We can also get the solution by graphing: we draw the lines corresponding to the equations in a coordinate system. The point of intersection gives us the solution, as pictured above!

• Decide which graph belongs to the system of equations.

  Hints

  You can check the intersection of the lines each time. The intersection point must satisfy both equations.

  You can write an equation like $2x+y=3$ in slope-intercept form by subtracting $2x$ from both sides: $y=-2x+3$.

  First, draw the $y$-intercept and then the slope; remember, it's rise over run!

  Lines corresponding to $y=mx$ pass through the origin.

  Solution

  To draw the lines corresponding to a system of equations, we first write each equation in a slope-intercept form: $y=mx+b$:

  $\begin{array}{rcl} x+y&=&6\\ y&=&2x \end{array}$

  The second equation is already given in slope-intercept form. Dividing both sides by $x$ in this equation leads to $y=-x+6$. So we have to look for two lines, one with a $y$-intercept of zero and the other with a $y$-intercept of $6$. The solution is given by $x=2$ and $y=4$.

  Now we look at the next system of equations:

  $\begin{array}{rcl} x-y&=&0\\ -x+3y&=&6 \end{array}$

  The first equation is equivalent to $y=x$ and the second to $y=\frac13x+2$. The $y$-intercepts are zero and $2$. The solution is given by $x=3$ and $y=3$.

  The next one:

  $\begin{array}{rcl} 2x+3y&=&12\\ 5x+3y&=&21 \end{array}$

  The first equation can be written in slope-intercept form: $y=-\frac23x+4$. The second as well can be written in slope-intercept form: $y=-\frac53x+7$. The $y$-intercepts are $4$ and $7$ and the solution is $x=3$ and $y=2$.

  And for the last example:

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

  Rewriting both equations in slope-intercept form gives us $y=-2x+6$ and $y=x+3$, with $y$-intercepts $6$ and $3$. The solution is given by $x=1$ and $y=4$.