Solve Linear Systems - International Brotherhood of Electrical Workers Exam IBEW

1. Two apprentices, Alex and Ben, are paid a total of $560 for a job. Alex worked 5 more hours than Ben, and their combined hours totaled 35. If they are paid at the same hourly rate, how much did Alex earn?
   Answer: $320 This is a multi-step problem. First, find the hours worked by each apprentice. Let $A$ be Alex's hours and $B$ be Ben's hours. 1) $A + B = 35$ 2) $A = B + 5$ Substitute the second equation into the first: $(B + 5) + B = 35$ $2B + 5 = 35$ $2B = 30 \rightarrow B = 15$ hours. Alex worked $A = 15 + 5 = 20$ hours. Next, find the hourly rate. The total pay for 35 hours is $560. Rate = Total Pay / Total Hours = $560 / 35 hours. To simplify $560/35$, we can divide both by 7: $80/5 = 16$. The hourly rate is $16. Finally, calculate Alex's earnings: Alex's earnings = Alex's hours $\times$ hourly rate = $20 \times \$16 = \$320$.
2. An electrician needs to create 500 feet of a wire bundle from two different spools of wire. The wire from spool A weighs 0.2 pounds per foot, and the wire from spool B weighs 0.3 pounds per foot. The final bundle must weigh exactly 120 pounds. How many feet of wire from spool A must be used?
   Answer: 300 feet Let $a$ be the length of wire from spool A and $b$ be the length from spool B. We can set up a system of two equations: 1) Total length: $a + b = 500$ 2) Total weight: $0.2a + 0.3b = 120$ We want to find the value of $a$. We can use substitution. From the first equation, $b = 500 - a$. Substitute this into the second equation: $0.2a + 0.3(500 - a) = 120$ $0.2a + 150 - 0.3a = 120$ Combine the $a$ terms: $-0.1a + 150 = 120$ Subtract 150 from both sides: $-0.1a = -30$ Divide by -0.1 (which is the same as multiplying by -10): $a = 300$ 300 feet of wire from spool A must be used.
3. The total length of two pieces of conduit is 18 feet. If three times the length of the shorter piece is 6 feet more than the length of the longer piece, what is the length of the longer piece?
   Answer: 12 feet Let $L$ be the length of the longer piece and $S$ be the length of the shorter piece. From the problem statement, we can create two equations: 1) The total length is 18 feet: $L + S = 18$ 2) 'Three times the shorter piece ($3S$) is (=) 6 feet more than the longer piece ($L+6$)': $3S = L + 6$ Now we solve the system. From equation (1), we can write $S = 18 - L$. Substitute this into equation (2): $3(18 - L) = L + 6$ $54 - 3L = L + 6$ $54 - 6 = L + 3L$ $48 = 4L$ $L = 12$ The length of the longer piece is 12 feet.
4. What is the $x$-coordinate of the solution to the system: $\frac{2}{3}x - y = 4$ and $x + \frac{1}{2}y = 1$?
   Answer: 9/4 To solve this system, it is easiest to first eliminate the $y$ variable. We can modify the second equation so the $y$ coefficient is the opposite of the $y$ coefficient in the first equation. The first equation has a $-y$ term. Multiply the second equation by 2: $2(x + \frac{1}{2}y) = 2(1) \rightarrow 2x + y = 2$ Now we have the system: 1) $\frac{2}{3}x - y = 4$ 2) $2x + y = 2$ Add the two equations together: $(\frac{2}{3}x - y) + (2x + y) = 4 + 2$ $\frac{2}{3}x + 2x = 6$ To add the $x$ terms, find a common denominator: $2x = \frac{6}{3}x$. $\frac{2}{3}x + \frac{6}{3}x = 6$ $\frac{8}{3}x = 6$ To solve for $x$, multiply both sides by the reciprocal of $\frac{8}{3}$, which is $\frac{3}{8}$. $x = 6 \times \frac{3}{8} = \frac{18}{8} = \frac{9}{4}$.

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