## High School Math Problems

*A change purse has 100 nickels and dimes. The total value of the coins is $7. How many coins of each type does the purse contain?*

**Problem 1:**If the number of nickels is N and the number of dimes is D, then

**(the 5, 10 and 700 representing the number of cents)**

5N + 10D = 700

However,

**(the number of nickels plus the number of dimes equals 100)**

N + D = 100

So, solving for N for both equations, we get as the result

**and**

N = – 2D + 140

**Thus,**

N = 100 – D

**Since the number of dimes equals 40, then the number of nickels must equal 60. Does this prove to be true?**

– 2D + 140 = 100 – D

D = 40

**————————-**

60 (5) + 40 (10) = 300 plus 400 = 700 cents. *Correct Answer*

*How Do You Multiply Fractions?*

**High School Math Problem 2:**Example Problem: 1/6 x 3-3/4 [one-sixth times three and three-quarters]

Perhaps the easiest way to multiply fractions is to first convert the numbers entirely into their fraction equivalents. Thus,

1/6 x 3-3/4 becomes 1/6 times 3 converted to fourths (12/4) added to the 3 fourths (for a total of 15/4). Then rewrite the problem as 1/6 x 15/4 = What?

Multiply the two top numbers and the two bottom numbers. 1 x 15 = 15, and 6 x 4 = 24.

Then put the resultant top number on top of the resultant bottom number. That is:

**Now both the top and bottom numbers can be divided by 3. So doing that to the 15/24 gives us 5/8.**

15/24

**————————-**

1/6 x 3-3/4 = 5/8 *Correct* *Answer*

*Solve by Factoring: x² – 6x – 27 = 0*

**High School Math Problem 3:**Look at the coefficient in front of the x (6) and the pure number (– 27). Clearly a factor of 3 is involved. 27 divides by 9, and 6 = 9 – 3. So making an intelligent first guess,

**Let’s perform a check to see if this is correct. Multiply the first term in the left expression by the first term in the second expression. That gives x times x = x². Now multiply the first term in the left expression by the second term in the right expression. That gives x times – 9 = – 9x. Next, multiply the second term in the left expression times the first term in the right expression. That gives 3 times x = 3x. Finally, multiply the second term in the left expression times the second term in the right expression. That gives 3 times – 9 = – 27.**

(x + 3) (x – 9) = 0 *Answer*

Putting these all together gives:

**That is the same as,**

x² – 9x + 3x – 27

**————————-**

x² – 6x – 27 *Correct*

*The sum of two numbers is 400. If the first number is decreased by 20% and the second number is decreased by 15%, then the sum is 68 less. Find the new numbers (the numbers after decreasing).*

**High School Math Problem 4:**The two original numbers (we’ll call them x and y) equal 400 total. Thus,

**But if we reduce x by 20 percent and y by 15 percent, and the sum is 68 less than the 400 (or 332), we write,**

x + y = 400 [Equation 1]

**We want to solve these two equations for x. So we multiply the second equation by 1.25, which gives,**

0.8x + 0.85y = 332

**Subtracting the first very first equation from the second, gives us,**

x + 1.0625y = 415 [Equation 2]

**This tells us the first value of y was 240. That means the first x was 160.**

0.0625y = 15

The new values are reduced by the cited percentages, and we wind up with,

**New Y = 204 and New X = 128** *Correct Answer*

*Solve Two Equations for x and y*

**High School Math Problem 5:**Use the following two equations to solve for the variables x and y.

**In both instances, we will attempt to “solve” for y. That means we will put y on the left of both equations and move everything else to the right.**

2x + 2y = 6 [Equation 1]

– 3x + 5y = – 33 [Equation 2]

Signs change (plus to minus and minus to plus) when you cross from one side of the equals symbol to the other.

Equation 1 becomes:

**This simplifies (by dividing everything by 2) to**

2y = 6 – 2x

**Equation 2 becomes**

y = 3 – x (New Equation 1)

**Dividing by 5 gives,**

5y = – 33 + 3x

**Now we can combine or solve the two equations. Since both say y equals something, we can join the equations. If a = b and b = c, then a = c. That’s the principal.**

y = – 33/5 + (3/5)x (New Equation 2)

So,

**This becomes (again, by moving across the equals sign)**

3 – x = – 33/5 + (3/5)x

**————————-**

– x – (3/5)x = – 33/5 – 15/5

(– 8/5)x = – 48/5

– 8x = – 48

x = 6 *Correct* *Answer*

*Find the slope intercept form of the line passing through the point (– 1, 5) and parallel to the line – 6x – 7y = – 3.*

**High School Math Problem 6:**The line given is rewritten (in slope-intercept form, y = mx + b) as

**Thus the slope is,**

y = – 6/7x + 3/7

**Now a line is parallel to another line if its slope is the same as the other line. So,**

m = – 6/7

y = – 6/7 x + b is the formula for the new line, with the intercept being b because we have not yet solved for it. We now do so, inserting the value of x and the value of y from the point. We get,

**The parallel line is represented by the equation:**

5 = – 6/7 (– 1) + b

b = 29/7

**y = – 6/7x + 29/7 ** *Correct Answer*

*Write an Equation of the Line Parallel to the Line 3x ‒ 2y = 8*

**High School Math Problem 7:**The general equation for a line is usually written,

y = mx + b

The variable m refers to the tilt or “slope” of the line. The letter b refers to where the line crosses the y-coordinate axis.To make the problem intelligible, the equation 3x ‒ 2y = 8 should be rewritten in the slope-intercept form. Keeping y on the left and moving x to the right, we obtain,

**Multiplying by ‒ 1, we get**

‒ 2y = ‒ 3x + 8

**Dividing by 2 yields**

2y = 3x ‒ 8

**We will call this Equation 1. We hold onto this and engage in the same process for the second equation.**

y = 3/2 x ‒ 4

**Moving x to the right, we get**

2y + 3x = ‒ 4

**Then dividing by two, we get,**

2y = ‒ 3x ‒ 4

**We will call this Equation 2.**

y = ‒ 3/2 x ‒ 2

An equation is parallel if the tilt or slope, the m value, exactly matches for both equations.

This high school math problem calls for us to use the intercept (b value, Equation 2), which is ‒ 2. It wants us to use the slope or m value of Equation 1, which is 3/2.

Our parallel line equation becomes,

**Now admit it. High school math problems can be both fun and entertaining.**

y = 3/2x ‒ 2 *Correct Answer*

**References:**

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Math is one of those subjects that even the best will make errors. In case you’re wondering, I’m not the best. However, I would greatly appreciate your notifying me should you spot any error, however small. Thank you in advance!