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# Tutor profile: Jamie S.

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Jamie S.
Math Teacher (for 10 years), Math tutor (for 1 year)
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## Questions

### Subject:Algebra

TutorMe
Question:

Cell phone carrier A offers a plan for \$40/month plus \$0.25/MB of data used. Cell Phone carrier B offers a plan for \$30/month plus \$0.30/MB used. How can y/ou use a system of equations to figure out which cell phone plan is the best deal? When is Plan A a better deal? When is Plan B a better deal?

Inactive
Jamie S.

You can use a system of equations by writing an algebraic equation to represent each carrier's plan. Plan A: C=40+0.25m ; Plan B: C=30+0.3m ; where C = monthly cost and m = MB of data used. Then you can solve the system with a graph or by setting the equations equal to each other (substitution), to determine how many MB you would need to use so that the plans would have an equal monthly cost. Then you would need to determine if you are going to use more or less MB than that to see which plan will work best for you. If you use more than 200 MB of data each month, then Plan A will be a better deal for you. If you use less than 200 MB of data each month then Plan B will be a better deal.

### Subject:Basic Math

TutorMe
Question:

Use factoring to solve this quadratic equation. x^2 + 2x - 10 = 5

Inactive
Jamie S.

Start by getting the equation to equal 0. To do this you would subtract 5 from each side. x^2 + 2x - 10 - 5 = 5 - 5 x^2 +2x -15 = 0 Now factor the equation. You are looking for two numbers that add to 2 and multiply to -15. You know you are looking for one positive and one negative number since the multiply to a negative. The two numbers are 5 and -3. You will use these two numbers in your two binomials that you factor. (x+5) (x-3) = 0 Since you have two factors being multiplied to 0, you know that one of them has to equal 0. Set each factor to equal 0 and solve each new equation. x+5=0 and x-3=0 x= -5 and x=3 These are your two solutions to the original quadratic equation. To check your answers, plug each number in (separately) the original equation to see that it makes the equation true. x^2 + 2x - 10 = 5 (-5)^2 + 2(-5) - 10 = 5 25 + (-10) - 10 = 5 15 - 10 = 5 5 = 5 (true) x^2 + 2x - 10 = 5 (3)^2 + 2(3) - 10 = 5 9 + 6 - 10 = 5 15 - 10 = 5 5 = 5 (true)

### Subject:Pre-Algebra

TutorMe
Question:

Without graphing or solving, how can you determine whether the following system of equations will have one solution, no solutions, or infinitely many solutions? y = -3x + 4 6x + 2y = 8

Inactive
Jamie S.

To determine the number of solutions with out graphing or solving you need to determine if these two linear equations are intersecting lines, parallel lines, or the same line. If they are intersecting lines, they will have two different slopes. If they are parallel lines they will have the same slope, but different y intercepts. If they are the same line they will have the same slope and the same y intercept. To find the slope and y intercept it is easiest to put each equation in slope-intercept form (y=mx + b; where m is the slope and b is the y intercept.) The first equation is already in this form so we just need to change the second equation by solving for y. 6x + 2y = 8 6x -6x +2y = 8 - 6x (subtract 6x from bothh sides) 2y = -6x + 8 (Simplify) 2y/2 = -6x/2 + 8/2 (divide each term by 2, to isolate y) y = -3x + 4 Since the two equations have the same slope and the same y intercept, there will be infinitely many solutions to this system of equations.

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