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# Live Math Help

Math can be a tough subject to learn in school. Live math help takes care of this by offering one-on-one help with an expert online tutor. Live math help comes with personalized tutoring for students to solve easy and complex math problems. Regular sessions with a tutor also help improve concept understanding and grades. Using a live math help service, students can study from the comfort of home using the Internet. Live math help sites quickly and instantly connect students with experienced math tutors who are subject experts and have extensive tutoring experience.

## Solve Math Problems

Live math help sites also come with extensive practice sheets and exercises on each topic to make sure a student confident about what they are learning. This is a great option for students who are looking for serious help with the subject. With the convenience of not having to drive to a learning center and the additional advantage of getting to study one-on-one with a tutor, live math help meets every student's learning needs. Extensive solved examples and worksheets, live math help sites are the best way to study the subject.

### Solved Examples

Question 1: The sum of two numbers is 35. One number is four less than the other number. Find the smaller number.
Solution:
Let two numbers be x and y.
and y < x (y is smaller than x)

Step 1:
Sum of two numbers = 35

=> x + y = 35                             ..............................(1)

and one number is four less than the other number

=> y = x - 4, because y < x

Step 2:
Put y = x - 4 in equation (1)

=> x + x - 4 = 35

=> 2x - 4 = 35

=> 2x - 4 + 4 = 35 + 4

=> 2x = 39

Divide each side by 2

=> x = $\frac{39}{2}$, put in y = x - 4

=> y = $\frac{39}{2}$ - 4

=> y = $\frac{39 - 8}{2}$ = $\frac{31}{2}$

Hence the smaller number is $\frac{31}{2}$.

Question 2: The area of a rectangular field is 81 square meter. If the length of the field is 3 meter, find the breadth of the field.
Solution:
Area of the rectangular field = 81
Length of the field = 3
Let breadth of the field = x

[Area of the rectangle = Length * Breadth]

=> 81 = 3 * x

Divide each side by 3

=> $\frac{81}{3} = \frac{3x}{3}$

=> 27 = x

Hence, breadth of the rectangular field is 27 meter.

Question 3: Find the value of 'm' if y = 3 is the solution of the quadratic polynomial

5 + (m + 2)x - x2
Solution:
Given quadratic polynomial, 5 + (m + 2)x - x2

Let P(x) = 5 + (m + 2)x - x2

Since y = 3 is the solution of P(x)

=> P(3) = 0

=> P(3) = 5 + (m + 2)3 - 32 = 0

=> 5 + 3m + 6 - 9 = 0

=> 2 + 3m = 0

=> 3m = - 2

=> m = $\frac{- 2}{3}$. answer

## Live Math Homework Help

Live math homework help will not give students ready-made answers to their problems, but rather show students how to get to the answers themselves. Some sites offer regular math homework and assignment help to students. Math is one subject that students find tough to understand. The difficult problems, endless practice questions and equations certainly don't make it easy. You work one-to-one with your tutor in our online classroom on your specific homework problem until it's done. Live Homework Help provide the brief and distinct answers to the student problems.

### Solved Examples

Question 1: Charlus is 12 years younger than Dunela. The sum of their ages is 90. Find the Charlu's age.

Solution:
Let age of Charlus = x

and age of Dunela = y

Step 1:
Sum of their age = 90

=> x + y = 90                      ..............................(1)

Charlus is 12 years younger than Dunela

=> x = y - 12                      ...............................(2)

Step 2:
Put equation (2) in equation (1)

=> y - 12 + y = 90

=> 2y -12 = 90

=> 2y = 90 + 12 = 102

=> y = 51

Hence, Charlus is of 51.

Question 2: Simplify ($\sqrt{3}$ + $\sqrt{2}$)2
Solution:
($\sqrt{3}$ + $\sqrt{2}$)2  = ($\sqrt{3}$ )2 + ($\sqrt{2}$)2 + 2 * $\sqrt{3}$  * $\sqrt{2}$

[(a + b)2 = a2 + b2 + 2ab]

= 3 + 2 + 2$\sqrt{6}$

= 5 + 2$\sqrt{6}$

=> ($\sqrt{3}$ + $\sqrt{2}$)2 = 5 + 2$\sqrt{6}$. answer

Question 3: Solve the quadratic equation, x2 + 5x = 0
Solution:
x2 + 5x = 0

=> x(x + 5) = 0

=> either x = 0  or  x + 5 = 0

Since given equation is a quadratic equation, so x $\neq$ 0.

=> x + 5 = 0

=> x = - 5, is the answer.