50 challenging algebra problems (fully solved)

Solution to Problem 2Subtract y from both sides of the first equation.. Check the answer: Plug x ~ 3.281 into the original equation... Problem 5Directions: Determine the equation of a st

50 CHALLENGING ALGEBRA PROBLEMS

Zishka Publishing

ISBN: 978-1-941691-23-6

Textbooks > Math > Algebra

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Education > Math > Algebra

Problem 1

Directions: Solve for x in the equation below (Don’t use guess and check.)

• You can find the solution on the following page

Note that 32 + 2 = 34 and (32)(2) = 64.

The answer checks out.

Common mistake: It’s incorrect to rewrite

as x + 2 = 32 You can easily verify that x doesn’t equal 30 by plugging 30 infor x in the original equation with a calculator You can’t square each termindividually in algebra You can square both sides of the equation, but thenyou get a cross term when you f.o.i.l it out:

That’s why x + 2 = 32 is incorrect It should really be

Problem 2

Directions: Solve for x and y in the system of equations below.

• You can find the solution on the following page.

Solution to Problem 2

Subtract y from both sides of the first equation

x = 16 – ySubstitute 16 – y in place of x in the second equation

Add the left fractions by first making a common denominator Multiply

by and multiply by in order to make a commondenominator

To add fractions that share a common denominator, add their numerators.

Cross multiply Recall that becomes wz = xy when you crossmultiply.

(16)(3) = (1)(16y – y2)

48 = 16y – y2This is a quadratic equation First express the equation in standard form

y2 – 16y + 48 = 0Either factor this equation or apply the quadratic formula We will factor it

(y – 4)(y – 12) = 0

The two solutions are y = 4 and y = 12 Plug each solution into x = 16 – y.The two corresponding solutions are x = 12 and x = 4

Check the answers: Plug x = 4 and y = 12 into the original equations.

You will get the same thing using x = 12 and y = 4 The answers check out

Solution to Problem 3

Combine like terms: Add 3x3/4 to both sides and subtract 2x4/3 from bothsides

5x3/4 + 3x3/4 = 6x4/3 – 2x4/3

Note that 5x3/4 + 3x3/4 = 8x3/4 and 6x4/3 – 2x4/3 = 4x4/3 One way to see this is

to factor: For example, 5x3/4 + 3x3/4 = (5 + 3) x3/4 = 8x3/4

8x3/4 = 4x4/3Divide both sides of the equation by 4

2x3/4 = x4/3

Divide both sides of the equation by x3/4 (The problem states that x is

positive, so we don’t need to worry about dividing by zero.) Note that

Apply the rule

2 = x4/3–3/4Subtract fractions by making a common denominator Focus on the exponent

Since 4/3 – 3/4 equals 7/12, the previous equation can be expressed as:

If you enter 212/7 on a calculator, you will find that x ~ 3.281341424

Check the answer: Plug x ~ 3.281 into the original equation Use a

calculator

2x4/3 + 5x3/4 = 6x4/3 – 3x3/42(3.281)4/3 + 5(3.281)3/4 = 6(3.281)4/3 – 3(3.281)3/4

9.751 + 12.189 ~ 29.252 – 7.314Since this works out to 21.940 ~ 21.938, the answer checks out

Solution to Problem 4

Solving for x by applying algebraic operations to both sides of this equationwould be challenging However, it turns out that this problem is numericallysimple enough to figure out if you understand what the problem meansconceptually

The problem, xx = 256, is equivalent to asking, “What number can you raise

to the power of itself and obtain 256 as the result?”

Let’s try raising integers to the powers of themselves and see what happens

Problem 5

Directions: Determine the equation of a straight line that has a y-intercept of

1 and which is perpendicular to the line that is represented by the equationbelow

y = 2x + 3

• You can find the solution on the following page

Solution to Problem 5

Compare the given equation, y = 2x + 3, to the standard equation for a

straight line, which is y = mx + b The given line has a slope of m = 2 and ay-intercept of 3

We wish to find a new line that is perpendicular to the given line Recall thatlines are perpendicular if one slope is the negative of the reciprocal of theother That is,

where is the slope of the perpendicular line Plug in m = 2

The perpendicular line has a slope of negative one-half According to theproblem, the perpendicular line needs to have a y-intercept of 1, which wecan express as

into the general equation for a straight line in order to write an equation forthe perpendicular line

Check the answer: Here is a trick that you may learn in a trigonometry class

that can help you check the answer to questions like this using a calculator.(Be sure that the calculator is in degrees mode and not in radians mode.) Takethe inverse tangent of the slope, tan–1(m), in order to determine the angle thatthe line makes with the x-axis Be sure to use a tan–1 button or an arctangent(atan) button Don’t take 1/x of the tangent (because that would give youcotangent instead of the inverse tangent: in trigonometry, tan–1 doesn’t meanone over the tangent) According to a calculator, for the given line, tan–1(m) =tan–1(2) ~ 63.435°, while for the perpendicular line, tan–1 = tan–1(–1/2) ~ –26.565° One line is angled 63.435° above the x-axis, while the otherline is angled 26.565° below the x-axis (since the angle is negative) Addthese two angles to see that they make a right angle: 63.435° + 26.565° = 90°

If you have a graphing calculator, graph each line

Problem 6

Directions: Apply algebra to solve the following word problem.

A customer has ten coupons Each coupon offers a 10% discount off of thepurchase price Unlike many store coupons, these coupons don’t come withany restrictions (meaning that the customer may apply all ten coupons to thesame purchase) The customer buys a pair of boots with a retail price of $50.How much will the customer pay if the customer uses all ten coupons andthere is a sales tax of 8%? (Why aren’t the boots free?)

• You can find the solution on the following page

Solution to Problem 6

Before we apply algebra, let’s reason out why the boots aren’t free It’s

incorrect to multiply the number of coupons (10) by the percentage (10%).Why? Stores apply coupons one at a time, not all together

Here is what the program in the cash register does The retail price of theboots is $50 When the first coupon is applied, the program deducts 10%from the purchase price, subtracting $5 from $50 to make a new subtotal of

$45 When the second coupon is applied, the program deducts 10% off thenew purchase price, subtracting $4.50 from $45 to make a new subtotal of

$40.50 Although the first coupon was worth a $5 discount, the second

coupon was worth a $4.50 discount because the new purchase price was

lower Each coupon results in a smaller discount than the previous discount

as the purchase price is reduced That’s why the boots aren’t free

To carry out the algebra, we will multiply the retail price ($50) of the boots

by 0.9 ten times Instead of subtracting 10% from the purchase price ten

times (which would require determining the new purchase price after eachdiscount), we will multiply $50 by 90% ten times (since multiplying by 90%

is the same as reducing the price by 10%, as 100% – 10% = 90%) In decimalform, 90% equals 0.9 (divide by 100 to determine this) Multiplying by 0.9ten times equates to raising 0.9 to the power of 10 Finally, we will multiply

by 1.08 to add the 8% sales tax (add 8% to 100% to get 108%, then divide by

100 to turn 108% into a decimal) The customer will pay a total amount givenby:

problem says to apply algebra in the solution, but we’re just checking theanswer now.)

Solution to Problem 7

This is actually a quadratic equation One way to see this is to let y = x2

12x2 = 63 – 3x412y = 63 – 3y2

Add 3y2 to both sides and subtract 63 from both sides to put this in standardform

3y2 + 12y – 63 = 0It’s convenient to divide both sides of the equation by 3

y2 + 4y – 21 = 0Compare this to the general equation ay2 + by + c = 0 to determine that:

a = 1

b = 4

c = –21Plug these values into the quadratic formula (Alternatively, you could factorthe previous quadratic equation.)

Note that the two minus signs make a plus sign: –4(1)(–21) = (–4)(–21) =+84.

There are two possible answers for y:

We’re not finished yet because we need to solve for x Recall that y = x2.Squareroot both sides to get Why ±? That’s because

solution for x (since is imaginary) The two real answers are:

Check the answers: Plug into the original equation

Note that and

.12(3) = 63 – 3(9) = 36The answers check out

Solution to Problem 8

Substitute the top equation (x = yz) into the bottom equations

y = z(yz + 1)

z = (yz – 1)yMultiply the left-hand sides together and the right-hand sides together

yz = z(yz + 1)(yz – 1)yApply the f.o.i.l method: (a + b)(c + d) = ac + ad + bc + bd

yz = yz(y2z2 – yz + yz – 1)

yz = yz(y2z2 – 1)

Divide both sides of the equation by yz (we don’t need to worry aboutdividing by zero since the problem specified positive solutions) Note thatyz/yz = 1

1 = y2z2 – 1Add 1 to both sides of the equation to get 2 = y2z2 Squareroot both sides toget Divide both sides of the equation by z to get

Plug this expression for y into the second equation from thebeginning of this solution

Multiply both sides of the equation by z to get Squareroot both sides.

Plug this expression into the equation

This is the same as:

Plug y and z into the equation x = yz The ’s cancel out

Check the answers: Plug x, y, and z into the original equations.

The answers check out (within reasonable rounding tolerance).

Problem 9

Directions: Solve for x in the equation below (Don’t use guess and check.)

• You can find the solution on the following page

Solution to Problem 9

Compare the wording of this problem to the wording of Problem 4 Noticethat this problem specifically states to “solve for x.” Unlike Problem 4, guessand check would not be satisfactory based on this difference in the

instructions (Most teachers don’t give full credit for guess and check

solutions when known techniques could be used to solve for an answer,

regardless of the wording.)

Recall that a squareroot can be expressed as a fractional exponent

Substitute x1/2 in place of in the given equation

xx1/2 = 27Note that x1 = x Plug this into the previous equation

x1x1/2 = 27Apply the rule that xmxn = xm+n

x1+1/2 = 27

Note that 1 + 1/2 = 1 + 0.5 = 1.5 = 3/2 (or make a common denominator towrite 1 + 1/2 = 2/2 + 1/2 = 3/2)

x3/2 = 27Raise both sides of the equation to the power of 2/3 Why? We will apply the

rule that (xm)n = xmn such that (xm)1/m = xm(1/m) = x1 = x.

(x3/2)2/3 = 272/3

x = 272/3

The power of 2/3 means to square 27 and take the cube root, in any order It

is simpler to take the cube root first, since

(because 33 = 27)

x = 272/3 = (271/3)2 = 32 = 9

Check the answer: Plug x = 9 into the original equation.

The answer checks out

Problem 10

Directions: Determine what comes next in the following pattern.

16x88x64x42x2

• You can find the solution on the following page

Solution to Problem 10

What can you do algebraically to each term in order to create the givenpattern? You can divide each term by 2x2 For example,

We applied the rule that

problem, divide the last term of the pattern by 2x2

The final answer is 1 Note that if your answer is 1x0 or x0, your answersimplifies to 1 because x0 = 1

Check the answer: Start with 1 and multiply by 2x2, and see if this

reproduces the same pattern in reverse

(1)( 2x2) = 2x2

2 2 4

(4x4)( 2x2) = 8x6(8x6)(2x2) = 16x8The answer checks out.

Problem 11

Directions: Apply algebra to solve the following problem.

Variables w, x, y, and z are related by the following equation

If x doubles, y triples, and z quadruples, by what factor does w change?

• You can find the solution on the following page

The coefficient (8) cancels Recall that the way to divide two fractions is tomultiply the first fraction by the reciprocal of the second fraction, as in

Now we will express the last sentence of the problem with algebra Thefollowing equations represent that x doubles, y triples, and z quadruples.

x2 = 2x1

y2 = 3y1

z2 = 4z1Plug these expressions into the previous equation

Note that (4z1)2 = 42z12 = 16z12 according to the rule (xy)m = xmym Also

note that x1, y1, and z1 all cancel out For example,

Multiply both sides by w1 to get

Thus, wwill be 3/8 of its initial value.

Check the answer: Try it out with numbers, such as x1 = 3, y1 = 4, and z1 =2

Since (reduce by dividing 9 and 24 both by 3),the answer checks out.

Problem 12

Directions: Solve for x in the equation below.

• You can find the solution on the following page

Note that z + 2 – z = 2 (since z – z = 0) and 6y/2 = 3y (since 6/2 = 3)

Note that – 3y + 3y = 0 Look at that: Two of the variables (y and z) canceledout There is just one variable left: x Now the problem is easy

Factor out the x We’re applying the rule ax – bx = (a – b)x Note that 52 =25.

30 – 25 = 5

Multiply both sides of the equation by 2 and divide both sides by 5.

Note that y and z are both indeterminate However, the problem only asks forx

Check the answer: Plug x = 2 into the original equation (y and z cancel out

again)

The answer checks out.

Problem 13

Directions: Solve for x in the equation below (Don’t use guess and check.)

• You can find the solution on the following page

Solution to Problem 13

Multiply both sides of the equation by x1/6

x3/2x1/6 = 32Apply the rule xmxn = xm+n

x3/2+1/6 = 32Add the fractions by making a common denominator

(Divide 10 and 6 both by 2 in order to reduce 10/6 to 5/3.) Since 3/2 + 1/6 =5/3, the previous equation becomes:

x5/3 = 32

Raise both sides of the equation to the power of 3/5 Why? We will apply therule that (xm)n = xmn such that (xm)1/m = xm(1/m) = x1 = x.

(x5/3)3/5 = 323/5

x = 323/5

The power of 3/5 means to both raise to the 3rd power and take the 5th root,

in any order We will take the 5th root first and then cube it

The answer checks out (within reasonable rounding tolerance).

Problem 14

Directions: Apply algebra to solve the following word problem.

A rectangle has a perimeter of 15 m and an area of 9 m2 Determine thelength and width of the rectangle

• You can find the solution on the following page

Solution to Problem 14

Express the given information in equations We will use x for length and yfor width The area (A) of a rectangle equals its length times its width:

A = xyThe perimeter (P) of a rectangle equals twice its length plus twice its width:

P = 2x + 2yPlug in the given numbers for the area (A = 9 m2) and perimeter (P = 15 m)

9 = xy

15 = 2x + 2yDivide both sides of the first equation by x

Substitute this expression for y into the other equation

Multiply both sides of the equation by x Note that 2xx = 2x2 and 18x/x = 18.

15x = 2x2 + 18This is a quadratic equation First express the equation in standard form.Subtract 2x2 and 18 from both sides of the equation It’s convenient to thenmultiply by –1

2x2 – 15x + 18 = 0Either factor this equation or apply the quadratic formula We will factor it

(2x – 3)(x – 6) = 0There are two possible solutions:

2x – 3 = 0or

x – 6 = 0

The two solutions are x = 3/2 = 1.5 m and x = 6 m Plug each solution into9/x = y The two corresponding solutions are y = 6 m and y = 1.5 m Eitherway, the rectangle has a length of 6 m and a width of 1.5 m

Check the answers: Plug x = 1.5 m and y = 6 m into the original equations.

A = xy = (1.5 m)(6 m) = 9 m2

P = 2x + 2y = 2(1.5 m) + 2(6 m)

P = 3 m + 12 m = 15 m