Part 2 Verbal Problems Systems of Equations

Algebra 2 Tutorial

Part 2 Verbal Problems Systems of Equations

Intro

This is the second part with problems 4-6.

Sample Problem

Problem 4) The Perimeter of a rectangle is 86. If the width W is doubled and the length L is increased by 8, the new rectangle Perimeter is 140. Find the original width and length.

Problem 5) A motel clerk counts his $ 1 and $ 10 bills at the end of the day. He has a total of 52 bills and the money adds up to $ 151. How many $ 1 and $ 10 bills are there?

Problem 6) A 35% Sulfuric Acid Solution is to be added to 200 ml of a 12% Sulfuric Acid Solution to make a 17% Sulfuric Acid Solution.
How much of the 35% Sulfuric Acid Solution is to be added?
What is the volume of the resulting 17% Sulfuric Acid Solution?

Solution

Problem 4) To solve, Use P = 2W + 2L so 2W + 2L = 86 (EQ1)
and the new Width = 2W; new Length = L+8 then Use Substitution.
Then 2 (2W) + 2 (L+8) = 140 so 4W + 2L +16 = 140
Then 4W + 2L = 124 (EQ2) then Use Elimination and Subtract EQ1 from EQ2.
You get 2W = 38 (124-86) so W = 19. Replace 19 for W in EQ1
2(19) + 2L = 86 and 2L = 86 – 38 = 48 so L = 24. Answer W=19; L=24

Problem 5) Organize the information to obtain your equations – then solve.
There are X number of 1 dollar bills and Y number of 10 dollar bills.
The value of 1 dollar bills is X and the value of 10 dollar bills is 10Y.
The number of bills is X + Y = 52 (EQ1) and the Total value is
X + 10 Y = 151 (EQ2). Use Elimination so Subtract EQ1 from EQ2 and you get
9Y = 99 so Y = 11 and X = 41.
Check: 41×1 + 11×10 = 151.

Problem 6) Organize the information to obtain your equations – then solve.
For the 0.35 concentration, Let X = Solution Volume so the Volume of Sulfuric Acid is therefore 0.35X. For the 0.12 concentration, the Solution Volume = 200 ml so the Volume of Sulfuric Acid is therefore 24 ml (0.12 x 200). For the 0.17 concentration, Let Y = Solution Volume so the Volume of Sulfuric Acid is therefore 0.17Y.

This gives you the two equations needed:
Y = X + 200 (EQ1) and 0.17Y = 0.35X + 24 (EQ2)
Substitute EQ1 into EQ2 to get
0.17 (X+200) = 0.35X + 24 and 0.17X + 34 = 0.35X + 24
so 0.17X + 10 = 0.35X and 10 = 0.18X So X = 10/0.18 = 55.56 ml; Y = 255.56 ml
which means that 55.56 ml of the 35% solution needs to be added and the volume of the resulting 17% solution is 255.56 ml.



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