Part a

Pick two numbers, call them and . For the first statement: “When you add them you get 8” we can write:

For the second statement: “when squared and subtracted you get 16”, we can write this as:

Part b

If you try letting and , then we can test this in the first equation:

This statement is true, so there is hope. But if we try the second statement:

This is a false statement, so and is a solution to the first equation, but not the second equation.

Part c

We can repeat the same procedure. Checking the first equation, we get:

Which is obviously a true statement. But the last one went wrong on the second statement. So for the second statement:

So these values actually make both equations true at the same time. So you can confidently tell the riddler that the values are 5 and 3 as Robbin guessed.

Solving the Problem

1. Choosing a variable

We will choose to eliminate y here, so we highlight its coefficients

2. Multiplying by a constant to remove y

You can multiply to top equation by the highlighted and the bottom equation by . This would make the coefficients of y in both equations the same (both would be ):

3. Subtract the equations

So now if we subtract the left hand sides and the right hand sides, the y-values should just cancel each other out as shown:

4. Solve for directly

Now you can easily solve for the value:

5. Solving for Y

You could have done the same thing to solve for . Instead, we can just substitute into either equation to get :

Now we make y the subject:

Checking the result

So we notice that there is only one solution, where and . You can try substituting these values into both equations to show that both of them are true:

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Part a

Assuming there were children at the event that each payed 8 dollars, and there were adults at the event that each payed 20 dollars. If the total was 300,000 dollars, then:

Part b

We know that if we ignore the 3000 organizers (), we would need to double the number of children (), to equal the number of adults:

Part c

By making the two substitutions, we can work in groups of 1000 adults and children, which makes the math much simpler. Starting with our equations:

Substituting and , we get:

Of course the benefit here is that we can divide out a common factor of 1000 to get our new simpler equations:

Part d

To solve for the number of children, we have to make the number of adults in both equations the same. To do this, we start with our initial expressions:

Then we multiply the bottom equation by 20, which will make the A values the same:

Now at this stage, we just subtract the two expressions and simplify:

So that means there were 5000 children at the event.

Part e

To find the number of adults, we can just use our simplest equation. We can use any equation that contains both A and C. So I’ll use:

Moving the to the right, we get:

Given that , we get:

So there were 13,000 adults at the event in total.

Part f

To show that these values () are correct, we can substitute them into both of the original equations. So for the first equation we get:

This is a true statement. Then for the next equation, we can test it as well:

This is also true. So this solution satisfies both solutions simultaneously.

Part g

Given that there were 5,000 children and 13,000 adults, there were 18,000 people at the event in total.

1. Choose an expression to isolate

By looking at both expressions, it will be easiest to get the in the second expression on its own, so we will do that.

2. Make the subject of the second expression

This is straightforward, we just move everything over to the other side:

3. Substitute into the first equation

We chose to substitute into the first equation since we isolated in the second equation. So starting with the second equation, we’ve highlighted the in red:

We replace for since they are equal:

4. Solve the expression

Now we want to find the y-values that make this true. It is a quadratic, so there will be more than one y-value. We start by expanding and simplifying:

We can divide all terms by 8 to get:

Now we factorize a to get:

Now this tells us that we have two possible y values:

5. Substitute y to find x

So remember that we found:

So when

We can now substitute to find the other value as well:

So we have two pairs of solutions here:

• and
  

• and
  

So as you can see, equations that need substitution to solve, can have more than one pair of solutions.

Part a

First, start by making the subject of the first equation:

Then substitute it into the second equation:

Then we can expand this expression and simplify:

Now at this stage, we can expand the left hand side:

Now to actually solve for , we should factorize. So we take out from the first two terms and from the second two:

Then we factor out a :

Now we have two equations to solve. Since either or to make the whole left hand side go to zero. So solving for the first term:

Part b

Using the second equation (), we can solve for :

Part c

Now if we want to compare the strength when . We can find the strength while twisting by using

Then after this, we can use find the strength while pulling as well:

So as we can see, that when the thickness was , the strength was 13. So the strength has gone down while twisting, but it has gone up while pulling.

Then repeating this calculation for the strength while twisting is given by:

Then we can also repeat this while pulling to get:

So in this case, our strength while twisting has gotten larger, but our strength while pulling has gotten smaller than 13.

Therefore, this is the optimal thickness, because increasing or decreasing it will make the strength decrease in either twist or pull.

Part d

You can not solve this problem with elimination, because the two equations are both nonlinear:

• For , you’ve multiplied the two variables by each-other

• When we expand , we will end up with a on the right

Therefore, it is not possible to solve this with elimination. This is why it is important to be familiar with both methods.

Part a

Since Bobby is three years younger than Sam, if you take three away from Bobby’s age, you’ll have Sam’s age:

Similarly, if we take five years from Bobby’s current age (), we will have half of sam’s current age , so:

Part b

To solve this by elimination, we should get all the variables on the left in alphabetical order, and all the constants on the right:

So for the top equation, we will subtract and add 3 to both sides, then for the bottom equation, we will multiply everything by 2:

Now we just swap the 10 and the in the second equation:

Now solving this by elimination is actually pretty easy, we can just subtract the left hand sides and right hand sides of the two equations, since and will just cancel:

This tells us that Bobby is just 7 years old.

Now to find sam’s age, you can use any expression you like. But I’m going to choose one where is the subject. So looking up, we had . So if we substitute , we will get:

That means that Sam is 4 years old.

Part c

Now we will repeat this calculation with substitution, we know that:

So at this stage, we can swap the in the second expression:

Now we can solve this as we normally would. Multiplying both sides by 2, we get:

Now we can expand the left hand side and solve for :

This is the same answer we got by elimination.

Part d

Both of the methods were of similar difficulty. Generally, elimination is usually easier because it avoids division. But remember that it only works for problems that are linear (no powers of x or y).

Before starting this question, you need to come up with an expression for the cost of each plan as Benjy increases his usage. Suppose we let be the cost of plan a, and is the cost of plan b. Let be the amount of data that Benjy used, then the totals would be:

Part a

We want the cost of both plans to be equal, so in that case we want:

So replacing and for their expressions, we get:

Now we can solve this as we usually would. Start by moving all variables to the left and all constants to the right:

Then we remove the negative signs from both sides and divide by 0.07:

So once he’s used 857Mb of data, both plans would have the same cost.

Part b

To test which plan is actually cheaper, we should just try substituting a value that is much larger than the 857Mb break even point. Suppose we try 1000Mb. Into plan a, we get:

So he would owe $180 if he chose plan a. Repeating this for plan b, we would get:

So if he had chosen plan b, then he would owe $190. So as we can see plan a would be the cheaper of the two.

Part a

So to get rid of the fractions, start with the three expressions:

Then multiply the top one by 3, the middle one by 6, and the bottom one by 4:

Part b

Looking at all three of these expressions, removing is easy. But for demonstration purposes, we will remove instead. So from the first two equations, we will multiply the top equation by 4, because :

Now at this stage, we will subtract the top equation from the bottom equation to get rid of x:

Then we can divide by to get:

Now for the next expression, we can use the second and third equations (as long as we don’t use the same two). But we have to make sure to remove x again:

So to do this, we multiply the bottom equation by 4 again:

Now at this stage, we will subtract both expressions. We will subtract the bottom expression from the top expression and solve:

Part c

Now that we have two equations in terms of and :

We can multiply the bottom equation by 7 to get rid of :

Now at this stage, we subtract the two equations and solve for :

Then dividing by , we get the final value:

Now at this stage, we can use and substitute :

Then we subtract from both sides:

Finally, we can use to solve for :

Now the densities of are 40, 30 and 80 kilograms per liter respectively. Therefore, the heaviest liquid is liquid z.