Mathematics & Statistics

Solution:
A mathematical expression is just that: a “phrase” which represents a value. An equation is created when two expressions are set equal to each other – therefore, the easiest way to differentiate an expression from an equation is simply seeing whether or not an equal sign is present. An equation is an identity if the equality holds for all possible values of the variables involved. A constant term has a value which cannot change (e.g., “5”), while a variable term can represent different values (e.g., with different choices of ).

This first statement is an equation since there is an equal sign, but it is not an identity since we can choose a value of that breaks the equality. For example, if , then the statement claims that , which is obviously not true. The term is variable, while the terms 3 and 5 are constant.

Solution:
This is an expression since no equal sign is present. By extension, it is not an identity, since identities must first be equations. Both terms are variable here.

Solution:
This is an expression, even though no variables are present. The only term, 45, is a constant.

Solution:
This is an expression. The term is variable, while is constant.

Solution:
This is also an expression since there is still no equal sign. Both terms are variable.

Solution:
This is an equation, but not an identity since the equality does not hold for all values of . Choose for example, which implies that . This is not true, and therefore the equation is not an identity. The term on the left hand side of this equation is variable, while the one on the right is constant.

Solution:
This is an expression. Both terms are variable.

Solution:
This is an equation, though again not an identity. Choose , which implies that . This is not true, and therefore the equation is not an identity. The terms and are variable, while 1 and 2 are constant.

Solution:
In general, the th root of a number can be expressed as . In this case, the simplified form is , where .

Solution:
Following the same rule as the previous question, rewrite this first as . Exponent rules dictate that in general, , and so this simplifies to , where .

Solution:
Following the same rules as the previous question, the expression first reduces to , and simplification of the exponent then gives the answer as , where .

Solution:
In general, . The answer is therefore , where .

Solution:
Rewrite the expression as . Using the same rule as the previous question, the expression reduces to , where .

Solution:
Rewrite the expression as . This then simplifies to , where .

Solution:
Assuming , this simply reduces to as the numerator and denominator share common factor . In this case, .

Solution:
Here, the numerator and denominator share common factor . Assuming , cancellation gives , which then simplifies to , where .

Solution:

Solution:

Solution:

Solution:
Since no value of is given, an exact value is not possible as an answer.

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Solution:
since the top and bottom share a common factor of which cancels (assuming ).

Solution:
.

Solution:
by the distributive property.

Solution:
Divide both sides by 0.2: . To verify, plug this value of back into the original question and make sure that the statement is true. In this case, the choice of gives us . Since this is true, we know our solution is correct.

To give you an idea of what it would look like if we somehow got the incorrect answer, pretend we solved the original equation and got . Then, in the verification step, we would have . This is obviously false, and so is not a valid solution to the equation.

Solution:
Divide both sides by . This has the effect of multiplying each side by : , which reduces to show that . We verify by plugging this value back into the original question: . This is true, and so we’ve verified our answer.

Solution:
Begin by subtracting 8 from each side, yielding . Division by -6 gives . Verify the solution by plugging this value of back into the original equation: . This is true, and so we’ve verified our answer.

Solution:
On the left-hand side, a factor of is common to each term in the numerator, and cancels with the found in the denominator assuming : . Simplification gives , and adding 6 to both sides gives . Verify the solution by substitution: . This is true, and so we’ve verified our answer.

Solution:
To isolate the , raise both sides to the power. This gives . We plug in this value of to verify the answer: . This is true, and so we’ve verified our answer.

Solution:
In a similar fashion as the previous problem, raise both sides to the power, giving . This is approximately 1.447. A calculator is suggested to verify this one: . Our solution was rounded to begin with, which creates the small discrepancy in the verification, but this is approximately correct.

Solution:
In earlier questions, was the only variable present, and “solving the equation” ended with a numerical solution. In cases like these, we must recall that “solving for ” is synonymous with isolating on one side of the equation. In this equation, divide both sides by : due to the cancellation on the left-hand side of the equation. We can still verify this solution via substitution for in the original equation: where the final step is possible due to the cancellation of ‘s on the left hand side. This statement is true, and so we’ve verified our answer.

Solution:
In a similar fashion as the previous problem, we simply aim to isolate on one side of the equation. Divide by on both sides: . Verify via substitution: , where the final step is possible due to the cancellation of ‘s on the right hand side. This statement is true, and so we’ve verified our answer.

Solution:
This is an expression, so we are instructed to find the value. Plugging in , we have .

Solution:
This is also an expression, so we plug in and simplify: . This can simplify down to since both the numerator and denominator are divisible by 2.

Solution:
This is an equation, so we are instructed to solve for . The first step is still to plug in , giving

Solution:
This is also an equation, though there are two common trains of thought when it comes to solving it. Both start by plugging in the known value of , giving .
In the first train of thought, consider expanding the left hand side:

From here, we can use the quadratic formula to find two solutions: . Try them both and notice that plugging in both of these values into the original equation makes the statement true.
In the second train of thought, note that you can take the square root of both sides of the equation, giving . This is trickier! Keep in mind that the symbol, meaning “plus or minus”, means that the 3 here could be positive or negative. This is because .

Solution:
This can be rewritten as , which can be FOILed (recall this tells you the order in which to multiply terms – First, Outer, Inner, Last). This gives .

Solution:
Following the same logic as the last problem, the multiplication gives .

Solution:
Slope (commonly denoted by the symbol ) can be found via the familiar phrase “rise over run”. More accurately, we are concerned with the change in rise over the change in run . In this case, this comes out to .

Solution:
Following the same logic as the previous problem, we have .

Solution:
This represents the general case: .

Solution:
There are two lanes of thought here, corresponding to the two different generic forms of a line: slope-intercept form and point-slope form .

In slope-intercept form, the slope and y-intercept of the line need to be known (unsurprisingly). We calculated the slope between these points above: . Separately, when , we know that since is given as a point on this line. Therefore, the y-intercept, commonly denoted with the symbol , is 5. The equation of the line is therefore .

In point-slope form, the slope and any one point on the line must be known (unsurprisingly). Choose , for example: then which reduces to . Adding 5 to both sides gives the same answer as before: .

Point-slope form is generally more helpful, as needing any point on the line is a less strict requirement than knowing the y-intercept. The slope is needed in either case.

Solution:
We don’t know the y-intercept here, so use point-slope form with either of the points provided. For the purpose of this example, we’ll use ; recall from the earlier question that the slope between these points is -14. The equation of the line is . Adding 11 to both sides gives . (Note that from this manipulation, we can actually determine the y-intercept to be 39, as the final simplified answer is in slope-intercept form with . This isn’t obvious from the information given initially.)

Solution:
Use point-slope form again. For slope, recall that , and note that either point can be plugged into the point-slope formula. Both and are correct.

Solution:
The first row, consisting of the -values, can be thought of as function inputs and these values will be the first coordinates of the ordered pairs we’re looking for. The -values in the second row can be thought of as the function outputs corresponding to the respective inputs, and these will be the second coordinates of the ordered pairs. The answers are:

Solution:
From left-to-right, the points are

Solution:
The area of a rectangle of length and width is simply . In this case, the area is 45 .

Solution:
The area of a square with side length is simply . In this case, the area is 49 .

Solution:
A triangle with base and height has area . In this case, the area is 110 .

Solution:
The area of a circle of radius is . In this case, the area is , or approximately 50.265 .