Now, add the (modified) first equation and the second equation: -7x + 35y = x - 35y = = 0

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1 1. Solve the system of equations. I m going to use elimination, by multiplying the first equation by 7: 7(-x + 5y = 2) -7x + 35y = 14 D Now, add the (modified) first equation and the second equation: -7x + 35y = x - 35y = = 0 This results in a TRUE statement (0 = 0), which indicates that the system has infinitely many solutions. In other words, the two equations are actually the SAME LINE, just written differently. Notice that if you multiply the first equation by -7, you end up with the second equation. 2. Set up a system of equations. There are two unknowns: how much of the cheaper candy, and how much of the more expensive candy. Therefore, set up variables: E x = lbs. of cheaper candy y = lbs. of expensive candy The candy mixed together must weigh 30 lbs, therefore: x + y = 30 (equation 1) The resulting mixture must be worth $1.98/lb, and for 30 pounds of it, that would be: 1.98lbs * 30$/lb = $ Therefore, the cost per lb of each candy times it s weight (x or y) must add up to $ x y = (equation 2) Page 1 of 6

2 3. B For motion problems, I always set up a table: D (miles) R (mph) T (hours Boat 1 d 5 t Boat 2 60 d 10 t For each, D = RT, therefore write two equations: d = 5t 60 d = 10t rewrite as: d = 60 10t Equate the two expressions for d (i.e. substitution): 5t = 60 10t 15t = 60 t = 4 hours. 4. Solve. I will solve using elimination, so start by multiplying the first equation by -8: -8(x + 4y = 8) -8x 32y = -64 C Now add the (modified) first equation to the 2 nd equation: -8x 32y = x + 5y = y = 0 y = 0 Substitute y = 0 back into EITHER equation to find x: x + 4(0) = 8 x = 8 (8, 0) is the solution Page 2 of 6

3 5. Solve graphically. You CAN solve this graphically if you want, but it s easier to just use substitution or elimination. I will use elimination, by multiplying the 1 st equation by -6: Section 8.1 B 6(5x + y = 12) 30x 6y = 72 Now, add the (modified) 1 st equation to the 2 nd equation: 30x 6y = x + 6y = 9 27x = 81 x = 3 Substitute the value of x back into EITHER of the equations to solve for y: 5( 3) + y = y = 12 y = 3 solution is ( 3, 3) 6. Solve. I will use elimination by multiplying the 1 st equation by 3: 3(x + 6y = 41) 3x 18y = 123 B Now, add the (modified) 1 st equation to the 2 nd equation: 3x 18y = x + 5y = 32 Page 3 of 6 13y = 91 y = 7 Substitute the value of y back into EITHER of the equations to solve for x: x + 6(-7) = -41

4 x 42 = 41 x = 1 solution is (1, 7) 7. Start by making a table: C D (miles) R (mph) T (hours) Cruiser 20 r 5 t Power boat 40 r t For each, D = RT, therefore write two equations: 20 = (r 5)t 40 = rt 20 r 5 = t 40 r = t Solve each equation for t (above): Equate the two expressions for d (i.e. substitution): 20 r 5 = 40 r Cross-multiply: 20r = 40(r 5) 20r = 40r = 20r r = 10 mph So, the power boat is going 10 mph, and the cruiser is going 5 mph. Page 4 of 6

5 8. There are two different investments made that total $7000. Make a table showing the two investments. The amount invested at each interest rate is unknown, so let those be the variables x and y: A 10% interest 7% interest Total Amount x y 7000 invested ($) Interest rate Time 1 year 1 year Interest 0.1x 0.07y 580 The 1 st equation comes from the amount invested: x + y = 7000 The 2 nd equation comes from the total interest: 0.1x y = 580 You can clear the decimals in the second equation by multiplying the whole equation by 100: 100(0.1x y = 580) 10x + 7y = Now solve using either substitution or elimination. In this case I will use substitution, solving equation 1 for x: x = 7000 y Substitute this expression for x into the 2 nd equation: 10(7000 y) + 7y = y + 7y = = 3y y = 4000 Therefore, there was $4000 invested at 7%. Page 5 of 6

6 9. Solve. I m going to use elimination, so I ll start by multiplying both sides of the 1 st equation by 5: D 5(2x + 3y = -1) 10x + 15y = -5 Add the (modified) 1 st equation to the 2 nd equation: 10x + 15y = x 15y = 5 0 = 0 This results in a TRUE statement (0 = 0), which indicates that the system has infinitely many solutions. In other words, the two equations are actually the SAME LINE, just written differently. Notice that if you multiply the first equation by -5, you end up with the second equation. 10. A woman made a deposit of $200. If her deposit consisted of 68 bills, some of them one-dollar bills and the rest being five-dollar bills, how many one-dollar bills did she deposit? This is a very typical mixture problem, where there are two unknowns: # of one-dollar bills and # of five-dollar bills. Set up a table: D Page 6 of 6 $1 bills $5 bills Total Number x y 68 Total value of bills 1x = x 5y 200 Each row gives us an equation: x + y = 68 x + 5y = 200 I m going to subtract the 1 st equation from the 2 nd : x + 5y = 200 x + y = 68 4y = 132 y = 33 Therefore, x = = 35 one-dollar bills.