How to find the solution to an equation - ISEE Upper Level Quantitative Reasoning

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Question

Solve the following equation for y

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Answer

Solve the following equation for y

Begin by subtracting 7 from both sides

Next, divide both sides by 3 to get our final answer:

$y=\frac{42}{3}=14$

So, our answer is 14

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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?

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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?

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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?

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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_.

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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.

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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?

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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 :

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

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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 :

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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)

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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

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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

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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

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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.

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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)

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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)

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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

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

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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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Question

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

and are integers.

Which is greater?

(a) $\left \lfloor a\right \rfloor + \left \lfloor b \right \rfloor$

(b) $\left \lfloor a+ b \right \rfloor$

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Answer

(a) Since is an integer, $\left \lfloor a \right \rfloor = \left \lfloor x + 0.5 \right \rfloor = x$ .

Since is an integer, $\left \lfloor b \right \rfloor = \left \lfloor y + 0.5 \right \rfloor = y$ .

$\left \lfloor a\right \rfloor + \left \lfloor b \right \rfloor = x + y$

(b) By closure, is an integer, so

$\left \lfloor a+b \right \rfloor = \left \lfloor x + 0.5 +y+0.5 \right \rfloor = \left \lfloor x +y+1 \right \rfloor = x +y+1$ .

This makes (b) greater.

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Question

$x + 7 \sqrt{x} - 30 = 0$

Which is the greater quantity?

(a)

(b)

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Answer

Substitute $t = \sqrt {x}$ and, subsequently, $t^{2} = x$ :

$t^{2} + 7 t- 30 = 0$

Factor as , replacing the two question marks with integers whose product is and whose sum is . These integers are .

Break this up into two equations, replacing $\sqrt{x}$ for :

$\sqrt{x} = 3$

$\left ( \sqrt{x} \right )^{2} = 3^{2}$

or

$\sqrt{x} = -10$

This has no solution, since $\sqrt {x}$ must be nonnegative.

is the only solution, so (a) and (b) must be equal.

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