Modeling with Systems of Linear Equations

Let x represent the weight of the 40 % silver alloy in kilograms and y represent the weight of the 20 % silver alloy in kilograms. The sum of these two alloys should be equal to 5 kilograms. x+y=5 To create a second equation, we will first rewrite percentages as decimals. ccc Percent & → & Decimal [0.5em] 40 % &→ & 0.4 20 % &→ & 0.2 25 % &→ & 0.25 We want 5 kilograms of an alloy that contains 25 % silver. With this information in mind and using the above percentages, we can write the following equation. 0.4x+ 0.2y= 0.25(5) ⇕ 0.4x+0.2y=1.25 We can form our system of equations by combining the two equations. x+y=5 & (I) 0.4x+0.2y=1.25 & (II) Let's solve the system by using the Substitution Method. We will first isolate y in the first equation and substitute it into the second equation to find the value of x.

Next, we can substitute 54 for x in Equation (I), and solve for y.

Therefore, we need 154 kilograms of the alloy that is 40 % silver and 54 kilograms of the alloy that is 20 % silver. These values can also be expressed as decimal numbers. ccc Fraction & &Decimal [1em] 15/4 & → &3.75 [1em] 5/4 & → &1.25

To find the break even point for the scooter store, we will write a system of equations. Let s be the number of scooters sold and d be the amount of money from the sale of scooters. The store sells scooters for $100 each, so the income d can be represented by the following equation. d=100s Buying a scooter from the manufacturer costs the store $50. Additionally, the cost of operating the store each month is $2000. The expenses the store incurs each month is equal to the sum of these two costs. Since we are asked to find the break even point for the store, the sum must be equal to d. d=50s+2000 With these two equations, we can form a system of equations. d=100s & (I) d=50s+2000 & (II) If we solve this system of equations, we can find the monthly break even point for the store. To determine the number of scooters the store needs to sell, we solve for s. Let's use the Substitution Method.

The break even point is reached when 40 scooters are sold. Therefore, the store needs to sell at least 40 scooters per month to avoid loss.

Let p represent the number of pennies and q the number of quarters. We can write an equation knowing that Vincenzo wants to choose 12 coins. p+q=12 Vincenzo wants the coins to have a value of $ 2.28. Knowing that the value of a penny is $ 0.01 and that the value of a quarter is $ 0.25, we can write a second equation. 0.01p+ 0.25q= 2.28 Then we can combine these two equations we can form a system of equations. p+q=12 & (I) 0.01p+0.25q=2.28 & (II) Let's solve this system by using the Substitution Method. We will start by isolating p in Equation (I). Then, we will substitute its equivalent expression into Equation (II).

Now that we know the value of q, we can substitute it into Equation (I) to find the value of p.

Therefore, Vincenzo has to take 3 pennies and 9 quarters in order to make $ 2.28 from the 12 coins.

We know that the Panthers scored 68 points. We also know that each touchdown t is worth 7 points, after a one-point conversion, which can be expressed as 7t. Furthermore, each field goal f is worth 3 points, which we can write as 3f. With this information, we can write our first equation. 7t+4f=68 We are also told that the number of field goals was 1 more than twice the number of touchdowns. Let's write an equation to represent this statement. f=2t+1 The two equations we wrote form the system of linear equations that corresponds to choice B. cc A. t=2f+1 7t+3f=68 & B. f=2t+1 7t+3f=68 [1.75em] C. t=2f+1 3t+7f=68 &D. f=2t+1 3t+7f=68

To find the number of touchdowns and field goals that the Panthers scored last night, we will solve the system we wrote in Part A. Since f is already isolated in the first equation, we can use the Substitution Method.

The number of touchdowns the Panthers scored is 5. Now we can substitute this number for t into the first equation and simplify to determine the value of f.

The Panthers scored 11 field goals.

Let R be the rung number from the base of the ladder and t be the time in seconds. We know that LaShay climbs down the 30-rung ladder at a speed of 2 rungs per second. At the same time, Tiffaniqua climbs up an identical ladder at a speed of 1 rung per second.

We will write an equation for each person, then form a system of equations. LaShay starts at the 30^(th) rung and steps down 2 rungs each second. With this information, we can write our first equation. R = 30 - 2t Tiffaniqua starts from the base of the ladder and climbs up 1 rung every second. Let's write a second equation. R= 1t+0 ⇔ R=t If we combine these equations, we can form a linear system that we can solve by using the Substitution Method.

Tiffaniqua and LaShay will meet on the 10^(th) rung from the base.

To write a system of equations that represents the situation, we need to first assign variables to each item. We can say that the number of hardcover books is h and the number of paperback books is p. We know that Ramsha bought 10 books in an unknown combination of hardcovers and paperbacks. This can be illustrated by the following equation. h+p=10 We also know that each hardcover book costs $ 5 and that each paperback book costs $ 3. Furthermore, Ramsha spent a total of $38. This can be shown by a second equation. 5h+3p=38 The two equations above form a system of equations. h+p=10 & (I) 5h+3p=38 & (II) We can most easily use the Substitution Method to solve this system, since we can easily isolate either variable in the first equation. Let's begin by isolating h.

Ramsha bought 6 paperback books. We can find the value of h by substituting p=6 in the first equation. However, since we are only asked how many paperback books Ramsha bought, we do not need the value of h.