Question

2 examples please

Some word problems in this chapter involve one "easy equation", usually found by translating a simple relationship between x and y (like one is twice the other or that they add to a given number). The second linear equation often uses a "rate times quantity formula. This is true for • distance problems (rate.time = distance) • mixture problems (percent concentration amount of solution - amount of concentrate) money problems (value of one item. how many items - total value) Using a chart or drawings can help organize the different quantities. The actual equation you get depends on extra information; look for words like "total" to add, or equal to set them equal. Be careful with mixture problems since you will often need to multiply rate times quantity for the final mixture, not just add to a total Here is an example, then you need to solve a similar problem, showing 1. Make a chart or drawing, and state what your variables represent. Be clear and include any units. 2. Write the "easy equation". 3. Write the equation that involves multiplying rates and quantities, then usually add if appropriate (if know total) 4. Solve algebraically. I want you to use the Addition Method, since the Week 7 had you use substitution. 5. State the full solution in sentence form, including any units. EXAMPLE: A 20% acid solution is mixed with a 60% acid solution to make 600 liters of a 50% acid solution. How much of each solution was used? 1. X = amount of 20% solution, in liters - 600 Y= amount of 60% solution, in liters 60% 50% 20% 60% solution solution 50% solution Amount of solution y 600 in liters Amount of acid in 0.20x 0.60 liters 20% For this problem, I like to call the easy equation the "no spille equation. As you pour the two solutions together into the final container, if you don't spill the quantities should add up. 2. x + y - 600 The second equation notes that the amount of acid also adds up". You do not create more acid just by mading them together We need to multiply 50% times 600 L to get the acid in final mix. 3. 0.20% +0.60y = 0.50(600) Writing the system is now done. This is usually the hardest part of these problems! 4. To solve the system by substitution, I like to clear out the decimals from the second equation by multiplying by 10 (if there are fractions, multiply by the LCD). If we multiply the first equation by negative 2 and add, we can solve: - 2x - 2y = - 1200 2x + 6y3000 4y = 1800, divide by 4 to get y = 1800/4 = 450, and x = 600 - 450 = 150. S. SOLUTION: There are 150 L of the 20% acid solution and 450 L of the 60% acid solution. Now for the discussion question, solve the following problem using a system of equations similarly to how it was done above (not by some other method). Show your solving steps (use the addition method) and state your final answer in a sentence as done in the example above. Be sure to check your solution! $10,800 is divided into two investments, one earning 8% simple interest and the other earning 5% simple interest. After one year, they earn a combined interest of $795. How much was invested in each account?

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Let amount- invested at 8% ya Omount invested at 5% Buy 5°). simple interest Total amount Amount Simple indevent Invested LE 10,800 Aplevest received 0.05 y 795 0.08x Here. $.10800 divided into two investmenli su ) 2+4 = 10,800 0 and total interest received $795 . So 0.08x + 0.05 y = 795 Now, solve the system of equation a multiply the equation 6 by 8 and equation by 100 and subhrach them 0x8.- 2X100 - 88/184 = 86400 8x + 54 = 79500 34 - 6900 4:2300 10800 - 2300 8500 $ 8500 invested at: 86% Simple interest and $ 2300 invested at 5% simple interest-