Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Ashley L.
Tutor for more than ten years
Tutor Satisfaction Guarantee
SAT
TutorMe
Question:

If 1/3x + 1/5y = 2, what is 5x + 3y?

Ashley L.
Answer:

The SAT often asks that you recognizes the structure of an equation in order to be able to solve it. Notice that the question is not asking for the value of x or the value of y, but instead is asking for some combination of both variables. This is a good indication that you will need to manipulate the equation in some way, instead of working overtime to solve for x or solve for y. Ask yourself, what can we do the first equation, to make it look more like the expression we're solving for? If we multiply 1/3 and 1/5 by their least common denominator, we get 5 and 3, and the solution becomes easy. 1/3x + 1/5y = 2 (multiply both sides by 15) 5x + 3y = 30 30 is our solution.

ACT
TutorMe
Question:

If 2(x + 2)/4 = (3x - 7) / 8, what is the value of x?

Ashley L.
Answer:

There are two ways to solve this problem: The most straightforward is to cross-multiply, as you would for any proportion, then solve the equation. 8*2(x+2) = 4*(3x - 7) (simplify the left hand side) 16(x+2)=4(3x - 7) (distribute) 16x + 32 = 12x - 28 (subtract 32 from both sides) 16x = 12x - 60 (subtract 12 x from both sides) 4x = -60 (divide both sides by 4) x = -15 Another way, is to recognize that since the denominator of the fraction is half of the denominator on the right, by multiplying the numerator by 2, we are able to set the numerators equal to each other and solve. 2*2(x+2) = 3x - 7 (simplify the left hand side) 4(x+2) = 3x - 7 (distribute) 4x + 8 = 3x - 7 (subtract 3x from both sides) x + 8 = -7 (subtract 8 from both sides) x = -15

Algebra
TutorMe
Question:

A rectangular rug is placed on a rectangular floor. The length the floor is 4 feet greater than the width of the floor. The length of the rug is 2 feet less than the length of the floor. The width of the rug is 3 feet less than the length of the rug. Write an expression for the area of the floor that is not covered by the rug.

Ashley L.
Answer:

As with most story problems / word problems, the first thing to do is draw a picture. If we draw two rectangles, on inside the other, we can use go through the story line by line to label the length and width of the rug and the floor. First, we don't know anything about the width of the floor, so that will be our variable, w. From there, we can say that the length of the floor is 4 +w. The length of the rug is (4 + w) - 2, or w + 2. The width of the rug is (w + 2) - 3, or w -1. To find the area of the floor that is not covered by the rug, we must first find the area of the floor and the rug. To do this, we will multiply the length by the width. + For the floor, we multiply w(w + 4). Using the distributive property, we know the area of the floor is w^2 + 4w. There are no like terms there, so we are finished simplifying. For the rug, we multiply (w + 2)(w -1). Using the distributive property again, or what some teachers may call the FOIL method, we know that the area is of the rug is w^2 - w + 2w - 2. Combining the like terms, we get w^2 + w - 2. Finally we subtract the area of the rug from the area of the floor: (w^2 + 4w) - (w^2 + w - 2). Be SUPER careful with your subtraction here. w^2 - w^2 is 0. 4w - w is 3w. There is no constant term in the first expression, so we can use a 0 in its place. 0 - (-2) is +2. So our final area is 3w + 2.

Send a message explaining your
needs and Ashley will reply soon.
Contact Ashley
Ready now? Request a lesson.
Start Session
FAQs
What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Session" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.