How to find the solution to an equation - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

A company makes toy boats. Their monthly fixed costs are $1500. The variable costs are $50 per boat. They sell boats for $75 a piece. How many boats must be sold each month to break even?

Answer

The break-even point is where the costs equal the revenues

Fixed Costs + Variable Costs = Revenues

1500 + 50_x_ = 75_x_

Solving for x results in x = 60 boats sold each month to break even.

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Question

Sally sells cars for a living. She has a monthly salary of $1,000 and a commission of $500 for each car sold. How much money would she make if she sold seven cars in a month?

Answer

The commission she gets for selling seven cars is $500 * 7 = $3,500 and added to the salary of $1,000 yields $4,500 for the month.

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Question

Solve the following system of equations: x – y = 5 and 2_x_ + y = 4.

What is the sum of x and y?

Answer

Add the two equations to get 3_x_ = 9, so x = 3. Substitute the value of x into one of the equations to find the value of y; therefore x = 3 and y = –2, so their sum is 1.

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Question

If x = 1/3 and y = 1/2, find the value of 2_x_ + 3_y_.

Answer

Substitute the values of x and y into the given expression:

2(1/3) + 3(1/2)

= 2/3 + 3/2

= 4/6 + 9/6

= 13/6

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Question

Albert has thirteen bills in his wallet, each one a five-dollar bill or a ten-dollar bill. What is the fewest number of ten-dollar bills that he can have and have more than $100.

Answer

Let be the number of ten-dollar bills Albert has; then he has five-dollar bills.

He then has dollars in his wallet, which must be greater than $100. Set up and solve an inequality:

Therefore, the lowest whole number of ten-dollar bills that Albert can have is eight.

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Question

For what value(s) of is the expression $\frac{x^{2}-1}{x^{2}+9}$ undefined?

Answer

The expression is undefined for exactly those values of which yield a denominator of 0 - that is, for which

$x^{2} +9 =0$

However, for all real ,

$x^{2} \geq 0$ ,

and, subsequently,

$x^{2} +9 \geq 9 >0$

meaning the denominator is always positive. Therefore, the expression is defined for all real values of .

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Question

Solve for :

$x\left (3x+7 \right ) =20$

Answer

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

$x\left (3x+7 \right ) =20$

$x\left \cdot ; 3x+x\left \cdot; 7 =20$

$3x ^{2}+7x =20$

$3x ^{2}+7x -20=20 -20$

$3x ^{2}+7x -20=0$

Use the method to factor the quadratic expression $3x ^{2}+7x -20$ ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is . These integers are , so:

$\left (3x ^{2}-5x \right ) + \left ( 12x -20 \right ) =0$

$x \left (3x-5 \right ) +4 \left ( 3x -5 \right ) =0$

$\left (x +4 \right )\left (3x-5 \right )=0$

Set each expression equal to 0 and solve:

or

$x = \frac{5}{3} = 1\frac{2}{3}$

The solution set is $\left { -4, 1\frac{2}{3}\right }$ .

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Question

Answer

First, rewrite the quadratic equation in standard form by distributing the through the product on the left and collecting all of the terms on the left side:

$x \cdot 2x-x \cdot 9= 35$

$2x ^{2}-9x= 35$

$2x ^{2}-9x-35= 35 -35$

$2x ^{2}-9x-35= 0$

Use the method to factor the quadratic expression $2x ^{2}-9x-35$ ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

$\left (2x ^{2}-14x \right ) + \left ( 5x-35 \right ) = 0$

$2x \left (x-7 \right ) + 5\left (x-7 \right ) = 0$

$\left (2x+ 5 \right ) \left (x-7 \right ) = 0$

Set each expression equal to 0 and solve:

$x = - \frac{5}{2} = -2 \frac{1}{2}$

or

The solution set is $\left { -2 \frac{1}{2}, 7\right }$ .

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Question

Answer

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

$2x \cdot x+ 2x \cdot 4 -7\cdot x -7\cdot 4 = 27$

$2x^{2}+ 8x -7 x -28 = 27$

$2x^{2}+ x -28 = 27$

$2x^{2}+ x -28-27 = 27-27$

$2x^{2}+ x- 55 = 0$

Use the method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of . These numbers are , so we do the following:

$2x^{2}-10x+ 11x- 55 = 0$

$2x\left (x- 5 \right) + 11 \left ( x- 5 \right) = 0$

$\left (2x+ 11 \right) \left ( x- 5 \right) = 0$

Set each expression equal to 0 and solve:

$x = -5\frac{1}{2}$

or

The solution set is $\left { -5 \frac{1}{2}, 5\right }$ .

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Question

Solve for :

$x (x+ 5) = \left ( x+2\right )^{2}$

Answer

Expand both products, the left using distribution, the right using the binomial square pattern:

$x (x+ 5) = \left ( x+2\right )^{2}$

$x \cdot x+x \cdot 5= x^{2} + 2 \cdot 2 \cdot x +2^{2}\right )$

$x^{2}+5 x = x^{2} + 4x +4$

$x^{2} +5 x-x^{2} = x^{2} + 4x +4 -x^{2}$

Note that the quadratic terms can be eliminated, yielding a linear equation.

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Question

Solve for :

Answer

$4x + 2 \cdot x-2 \cdot7 = 2 \cdot 3x -2 \cdot5$

$\left (4 + 2 \right )x-14 = 6x -10$

This identically false statement alerts us to the fact that the original equation has no solution.

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Question

Which is the greater quantity?

(a)

(b)

Answer

If , then either or . Solve for in both equations:

or

Therefore, either (a) and (b) are equal or (b) is the greater quantity, but it cannot be determined with certainty.

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Question

Consider the line of the equation .

Which is the greater quantity?

(a) The -coordinate of the -intercept

(b) The -coordinate of the -intercept

Answer

(a) To find the -coordinate of the -intercept, substitute :

$4x + 5 \cdot 0 = 100$

$4x \div 4 = 100 \div 4$

(b) To find the -coordinate of the -intercept, substitute :

$4\cdot 0 + 5y = 100$

$5y \div 5 = 100\div 5$

(a) is the greater quantity.

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Question

Consider the line of the equation

Which is the greater quantity?

(a) The -coordinate of the -intercept

(b) The -coordinate of the -intercept

Answer

(a) To find the -coordinate of the -intercept, substitute :

$0.9x + 0.7 \cdot 0 = 100$

$0.9x \div 0.9 = 100 \div 0.9$

$x = 100 \div 0.9 = 1,000 \div 9 = 111 \frac {1}{9}$

(b) To find the -coordinate of the -intercept, substitute :

$0.9 \cdot 0 + 0.7y = 100$

$0.7y \div 0.7 = 100 \div 0.7$

$y =100 \div 0.7 = 1,000 \div 7 = 142\frac {6} {7}$

(b) is the greater quantity.

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Question

Which is the greater quantity?

(a)

(b) 0

Answer

can be rewritten as a compound statement:

or

Solve both:

or

Either way, , so (a) is the greater quantity

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Question

Consider the line of the equation .

Which is the greater quantity?

(a) The -coordinate of the -intercept.

(b) The -coordinate of the -intercept.

Answer

(a) To find the -coordinate of the -intercept, substitute :

$5x + 4 \cdot 0 = -200$

$5x\div 5= -200\div 5$

(b) To find the -coordinate of the -intercept, substitute :

$5 \cdot 0 + 4y = -200$

$4y \div 4= -200 \div 4$

Therefore (a) is the greater quantity.

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Question

Which is the greater quantity?

(a)

(b)

Answer

Each can be rewritten as a compound statement. Solve separately:

$45 - t =- 60 \textrm{ or }45 - t = 60$

or

Similarly:

$45 - u = -50 \textrm{ or } 45 - u = 50$

Therefore, it cannot be determined with certainty which of and is the greater.

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Question

Which is the greater quantity?

(a)

(b)

Answer

$\sqrt[3]{t^3 }=\sqrt[3]{ -125}$

$\sqrt[3]{t^3 }= -\sqrt[3]{ 125}$

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Question

Consider the line of the equation $\frac{1}{99}x + \frac{1}{101}y = 1$

Which is the greater quantity?

(a) The -coordinate of the -intercept

(b) The -coordinate of the -intercept

Answer

(a) To find the -coordinate of the -intercept, substitute :

$\frac{1}{99}x + \frac{1}{101}y = 1$

$\frac{1}{99}x + \frac{1}{101} \cdot 0 = 1$

$\frac{1}{99}x + 0 = 1$

$\frac{1}{99}x = 1$

$\frac{1}{99}x \cdot 99= 1\cdot 99$

(b) To find the -coordinate of the -intercept, substitute :

$\frac{1}{99}x + \frac{1}{101}y = 1$

$\frac{1}{99} \cdot 0 + \frac{1}{101}y = 1$

$0 + \frac{1}{101}y = 1$

$\frac{1}{101}y = 1$

$\frac{1}{101}y \cdot 101 = 1\cdot 101$

This makes (b) the greater quantity

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Question

$\left \lfloor N\right \rfloor$ refers to the greatest integer less than or equal to .

and are integers. Which is greater?

(a) $\left \lfloor x+ y\right \rfloor$

(b) $\left \lfloor x\right \rfloor + \left \lfloor y\right \rfloor$

Answer

If is an integer, then $\left \lfloor N\right \rfloor = N$ by definition.

Since , and, by closure, are all integers,

$\left \lfloor x+ y\right \rfloor = x + y$ and $\left \lfloor x\right \rfloor + \left \lfloor y\right \rfloor = x + y$ , making (a) and (b) equal.

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