Systems of Equations – Word Problems 07:19 minutes

Video Transcript

Transcript Systems of Equations – Word Problems

Jessica is volunteering at the local animal shelter. Today, she has to buy some food and medication for the dogs and cats at the shelter. Jessica needs to figure out how much she can spend on supplies for the dogs and cats. This is an example of solving word problems using systems of equations.

Systems of Equations

There are 16 cats and 24 dogs in the shelter. Jessica has to buy supplies for the animals at the shelter. We'll let 'c' equal the cost of supplies for each cat and 'd' equal the cost of supplies for each dog, where Dog supplies cost twice as much as cat supplies. If Jessica has $640 to spend, how much can she spend on cat and dog supplies, respectively?
Let's see what we have. 16 cats times the cost of cat supplies, 'c', plus 24 dogs times the cost of dog supplies, 'd', equals the total cost, $640. So our first equation is 16c + 24d = 640. So, we have two variables in one equation. The second piece of information given is that dog supplies cost twice as much as cat supplies. The second equation we have to use is d = 2c.

Using Substitution to solve a System of Equations

We can use substitution to solve this system of equations. Since the second equation is already solved for the variable 'd', we can substitute 'd' with 2c in the first equation. The result is one equation with one variable, which we can solve now. Multiplying 24 and 2c, gives us 48c. We can add 16c and 48c on the left side of the equation, since these are like terms. This leaves us with 64c. The last step is to solve the equation by dividing both sides of the equation by 64.
We now know Jessica can spend $10 per cat. We still don't know how much to spend on dog supplies. But we do know how much Jessica can spend per cat. Let's plug in 10 for 'c' into the second equation. It doesn't matter if we use the first or second equation for this step, but the second is much easier, don't you agree?
Now we can solve this equation for 'd' by multiplying. We've determined that Jessica can spend $10 per cat and $20 per dog. Like a good math student, we should check our work. The solution to a system of equations must satisfy both equations. We have to substitute 10 for 'c' and 20 for 'd' in both equations. Let's start with the first equation. On the left side of the equation, the first step is to multiply, left to right, according to PEMDAS. The next step is to add, and we can see each side of the equation equals 640, letting us know that our solution is correct. Let's check the second equation to see if it's also correct. On the right side of the equation, we just have to multiply, and we can see that each side of the equation equals 20. The solution is also true for the second equation.

But Jessica still thinks the animal supplies costs a pretty penny. Looking for a better deal, Jessica finds a new shop offering discount dog supplies. She can buy all of the supplies she needs for 12 dogs at $210. She wonders, if she has to spend less on dog supplies, for how many additional cats will there be money for supplies? Let's review the information we started with. We have an unknown number of cats, let this number equal 'x', with the cost of $10 for supplies for each cat. So we have 10x. We also have 24 dogs, and $640 to spend. This time we should use a new, second equation that reflects the store's special offer. Our second equation is 12 dogs times 'd', the cost of supplies for each dog, plus 0 cost for the number of cats times x, the number of cats, equals $210. We can use elimination to help solve this system of equations.
Notice what happens when the second equation is multiplied by -2, we get -24d = -420. Now we can eliminate the 24d in the first equation by adding the two equations together. Let's see what happens when we do that. We're left with 10x=220. Now all we have to do is divide both sides of the remaining equation by 10. Look at that!? Jessica can provide supplies for 22 cats with her budget. Since the shelter has 16 cats at the moment, she can provide supplies for 6 more cats.
Let's find out how much she spends with the special offer on the dog supplies. Either substitute 22 for 'x', or simply use the second equation since there is no 'x' anyway. The equation can be solved by dividing the number in front of 'd' in the equation. We know that Jessica spends $17.50 per dog at this store. Remember to check your work by substituting 22 for 'x' and 17.5 for 'd' into both equations. What a deal!

Solving a System of Equations graphically

In her pocket, Jessica finds $15 of her own. Jessica wants to buy some nice toys for her two favorite cats and her favorite dog with the money she saved. One doggy toy costs three times as much as one cat toy.
Let's do something a little different and solve this system of equations graphically. The first equation uses the information for 2 cats, 1 dog, and 15 dollars. The equation is 2x + y = 15. The second equation represents the fact that doggy toys are 3 times more expensive than cat toys. The equation is 3x = y. Each equation needs to be written in slope-intercept form to graph properly. For the first equation, subtracting 2x from both sides results in y = -2x + 15. The second equation is already in slope-intercept form.
Let's graph these lines on the coordinate plane. The solution to the system is the point of intersection for both lines. As we can see, the lines intersect at (3, 9) or x = 3 and y = 9. This means the cat toy is $3 and the dog toy is $9. Remember to do the check by substituting in 3 for 'x' and 9 for 'y' into both equations.

If only Jessica had known that the cats prefer the dog's toy . . .