Puzzle Problems

This week your student will work on solving linear equations. We can think of a balanced hanger as a metaphor for an equation. An equation says that the expressions on either side have equal value, just like a balanced hanger has equal weights on either side.

If we have a balanced hanger and add or remove the same amount of weight from each side, the result will still be in balance.

We can do this with equations as well: adding or subtracting the same amount from both sides of an equation keeps the sides equal to each other. For example, if $4x+20$ and $\text-6x +10$ have equal value, we can write an equation $4x+20=\text-6x+10$. We could add -10 to both sides of the equation or divide both sides of the equation by 2 and keep the sides equal to each other. Using these moves in systematic ways, we can find that $x=\text-1$ is a solution to this equation.

Here is a task to try with your student:

Elena and Noah work on the equation $\frac12 \left(x+4\right) = \text-10+2x$ together. Elena’s solution is $x=24$ and Noah’s solution is $x=\text-8$. Here is their work:

Do you agree with their solutions? Explain or show your reasoning.

Solution:

No, they both have errors in their solutions.

Elena multiplied both sides of the equation by 2 in her first step, but forgot to multiply the $2x$ by the 2. We can also check Elena’s answer by replacing $x$ with 24 in the original equation and seeing if the equation is true. $$\frac12 \left(x+4\right) =\text -10+2x$$ $$\frac12 \left(24+4\right) =\text -10+2(24)$$ $$\frac12 \left(28\right) = \text-10+48$$ $$14=38$$ Since 14 is not equal to 38, Elena’s answer is not correct.

Noah divided both sides by -3 in his last step, but wrote -8 instead of $8$ for $\text-24 \div \text-3$. We can also check Noah’s answer by replacing $x$ with -8 in the original equation and seeing if the equation is true. Noah’s answer is not correct.

Systems of Linear Equations

This week your student will work with systems of equations. A system of equations is a set of 2 (or more) equations where the letters represent the same values. For example, say Car A is traveling 75 miles per hour and passes a rest area. The distance in miles it has traveled from the rest area after $t$ hours is $d=75t$. Car B is traveling toward the rest area and its distance from the rest area at any time is $d=14-65t$. We can ask if there is ever a time when the distance of Car A from the rest area is the same as the distance of Car B from the rest area. If the answer is “yes,” then the solution will correspond to one point that is on both lines, such as the point $(0.1, 7.5)$ shown here. 0.1 hours after Car A passes the rest area, both cars will be 7.5 miles from the rest area.

We could also answer the question without using a graph. Since we are asking when the $d$ values for each car will be the same, we are asking for what $t$ value, if any, makes $75t=14-65t$ true. Solving this equation for $t$, we find that $t=0.1$ is a solution and at that time the cars are 7.5 miles away since $75t=75\boldcdot 0.1=7.5$. This finding matches the graph.

Here is a task to try with your student:

Lin and Diego are biking the same direction on the same path, but start at different times. Diego is riding at a constant speed of 18 miles per hour, so his distance traveled in miles can be represented by $d$ and the time he has traveled in hours by $t$, where $d=18t$. Lin started riding a quarter hour before Diego at a constant speed of 12 miles per hour, so her total distance traveled in miles can be represented by $d$, where $d=12\left(t+\frac14 \right)$. When will Lin and Diego meet?

Solution:

To find when Lin and Diego meet, that is, when they have traveled the same total distance, we can set the two equations equal to one another: $18t=12\left(t+\frac14 \right)$. Solving this equation for $t$, $$18t=12t+3$$ $$ 6t=3$$ $$t=\frac12$$ They meet after Diego rides for one half hour and Lin rides for three quarters of an hour. The distance they each travel before meeting is 9 miles, since $9=18 \boldcdot \frac12$. Another way to find a solution would be to graph both $d=18t$ and $d=12\left(t+\frac14 \right)$ on the same coordinate plane and interpret the point where these lines intersect.