Rae F.

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Pre-Algebra

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Question:

2(4X-5) - (7x + 4 + 1) = -14

Rae F.

Answer:

Remember order of operations: PEMDAS Parentheses Exponents Multiplication Division Addition Subtraction Parentheses: We add like terms 2(4X-5) - (7x + 4 + 1) = -14 2(4X-5) - (7x + 5) = -14 Exponents: There are none Multiplication: So if something is directly in front of parentheses we distribute (multiply it to everything inside) 2(4X -5) -1(7X + 5) if nothing is present place a 1 and distribute 2*4X + 2*(-5) -1*7X + -1*5 8X -10 -7X – 5 (Remember negative times positive makes negative) So now we have: 8X -10 – 7X – 5 = -14 Division: There is none Addition and Subtractions: We combine like terms 8X -10 – 7X – 5 = -14 1X – 10 – 5 = -14 1X – 10 – 5 = -14 Like terms with same signs you add and the answer will receive that sign. Like if you owe someone $10 then you owe them another $5 now you OWE them $15 1X – 15 = -14 Then we want to get X by itself so first we move everything over that is not attached to the X by doing to opposite So we ADD 15 to both sides (whatever you do to one side of the equation you have to do to the other 1X – 15 = -14 + 15 +15 1X = 1 If you have opposite signs you always subtract and then the larger numbers sign is the answers sign so 15 – 14 = 1. 15 was the bigger number and positive so the answer is positive. OR You owe the bank $14 but you put in $15 so now you have $1.

Basic Math

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Question:

2 1 – 1 4 3

Rae F.

Answer:

First separate whole numbers and fractions: Whole Fraction 2 1 4 - 1 3 Next, find a common denominator: this is a number that both 4 and 3 go into (a number that can be divided by both 4 and 3). I am going to use 12 First, we write what our new denominator will be next to each fraction Whole Fraction 2 1 __ 4 12 - 1 __ 3 12 Then we ask our selves for each fraction what did we have to multiply to get the new denominator. For the top, we multiplied by 3 and the bottom fraction we multiplied by 4. Whole Fraction 2 1 __ 4 x3 12 - 1 __ 3 x4 12 Then we must multiply the numerators by this same number. Whole Fraction 2 1 x3 __ 4 x3 12 - 1 x4 __ 3 x4 12 Whole Fraction 2 1 x3 3 4 x3 12 - 1 x4 4 3 x4 12 Then we combine the numerators: 3 – 4 Since we cannot take 4 away from 3 we must BORROW from the whole number Whole Fraction 2 1 1 x3 3 + 12 4 x3 12 12 - 1 x4 4 3 x4 12 We take away one of 2 making it 1. Then we add the 1 to the fraction in the form of a fraction using the common denominator. Anything over itself is 1: so we have all 12 pieces out of a pie that has 12 pieces so we eat the whole pie. So then we add the numerators: 3 + 12 to get out new number for the top fraction Whole Fraction 2 1 1 x3 3 + 12 = 15 4 x3 12 12 12 - 1 x4 4 3 x4 12 Now we can subtract: First the whole numbers 1 – 0 (we can say 0 because nothing is there in the while number section) than 15 -4 Whole Fraction 2 1 1 x3 3 + 12 = 15 4 x3 12 12 12 - 0 1 x4 4 3 x4 12 1 9 12 This can then be reduced because both 9 and 12 can be divided into 3 9 /3 Remember whatever you do to the numerator must also be done to the denominator 12 /3 9 /3 = 3 12 /3 4 So our answer is: 1 3 4

Algebra

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Question:

A total of 925 tickets were sold for a game for a total of $1,150. If adult tickets sold for $2.00 and children's tickets sold for $1.00, how many of each kind of ticket were sold?

Rae F.

Answer:

We know a total number of tickets sold was 925. So the number of adult tickets plus the number of children’s tickets sold adds to 925: Adult tickets + Children’s Tickets = 925 So we assign a variable to each: X (Adult tickets) + Y (Children’s Tickets) = 925 (total) X + Y = 925 We know the price of each ticket sold: $2.00 for Adult and $1.00 for children. To find the total price you must take the AMOUNT times by the PRICE sold to find the total. So we take $2.00 times X (Adult tickets) + $1.00 times Y (Children’s tickets) = $1,150 (Total amount made) 2X + 1Y = 1150 Now we have two equations: 2X + 1Y = 1550 X + Y = 925 There are two ways to solve: Elimination and Substitution Elimination: To eliminate you must get rid of one of the variables. This is done by adding the equations together. In order for something to eliminate, however, it has to cancel. This can be done by changing the values of 1 equations: you can multiply or divide by whatever value needed as long as it is done to BOTH sides of that equation 2X + 1Y = 1550 -1(1X + 1Y = 925) Remember if nothing is present in front of a variable it is equal to 1. Then distribute the -1 2X + 1Y = 1550 -1X – 1Y = -925 Then simply add the equations together 1X = 625 So X (Adult tickets) = 625 then we substitute this back into any equation to find the remaining Y (Children’s ticket) value X + Y = 925 625 + Y = 925 -625 -625 whatever is done to one side of an equation must be done to another Y = 300 Substitution: Another way to answer this is to take one equation and move everything to one side except 1 variable (does not matter which one) 2X + 1Y = 1550 X + Y = 925 X + Y = 925 -X -X Y = 925 – X Then this equation can be substituted into the remaining equation 2X + 1Y = 1550 2X + 1(925-X) = 1550 Then distribute 2X + 925 –X = 1550 Then solve by first adding like numbers and variables on each side 1X + 925 = 1550 Then move the 925 over by doing the opposite (it is added we subtract) - 925 -925 1X =625 Then substitute that into any equations to solve for Y X + Y = 925 625 + Y = 925 -625 -625 whatever is done to one side of an equation must be done to another Y = 300

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