Solving Math problems can be interesting with online learning sessions. Students can take these sessions based on their learning requirements and most importantly, they can schedule these sessions by staying at their home. It saves their time and gives personalized attention to students. They can work with proficient subject experts and solve tough Math problems in a virtual environment. It has been observed that most students face difficulties while solving Math problems and to tackle this situation, students need to revise each Math chapter in a thorough manner. Moreover, they can solve online Math questions, as well.

**Read it carefully**– Students need to read each Math problem carefully and consequently, they can solve it by using the right formula or concept. Reading as well as understanding the problem is necessary for a student in exams.

**Break the problem into parts**– Read and break the Math problem into two parts. Students need to solve each Math problem in a step-by-step manner by using the appropriate concepts.

**Change it into an equation**- It is important to convert the problem written in words into an equation. Then, students can solve it accurately.

**Always cross check**- Once you get the answer of Math problem, you should go back and recheck the entire steps. Sometimes, you may miss out some information that you need to include.

**Ask until it gets clear**- Math is all about concepts and hence, students need to understand Math concepts properly. Students should clear their doubts in each Math concept and then only, they can apply these concepts in a right manner. Moreover, students can opt for online Math sessions to solve each tough Math problem accurately.

To solve Math problems accurately and systematically, students need clear understanding of Math concepts. Solving Math problems is not an easy task, but students can make it easier by choosing online sessions with virtual tutors. TutorVista has a team of virtual Math tutors and they help students in understanding different Math problems in an accurate manner. Moreover, online tutors go through an extensive training before starting their sessions. Hence, they understand the student's learning problems and based on that they provide assistance to students.

Solving word problems is one of the tough problems in Math. It is added in the Math syllabus designed for different classes. It can happen in any Math topic like algebra, geometry and others. Students can choose online learning help for solving different Math problems. Free demo sessions are also available for this. Follow algebra word problems page offered by TutorVista as a reference.

Several expert Math tutors are associated with TutorVista and they guide students by providing detailed information about each Math topic. Moreover, by choosing online sessions with TutorVista, students can easily solve all math problems in an organized manner. Moreover, they can schedule as many sessions as they need to revise each topic. Apart from this, students can practice online Math questions to brush up their knowledge before tests.

**Solve problems in topics like:**

- Algebra
- Geometry
- Calculus
- Pre-Algebra
- Pre-calculus
- Trigonometry
- Discrete Mathematics
- Statistics

Students can take online learning help for solving algebra expressions, geometry problems, equations, probability, statistics, calculus and many more. TutorVista provides suitable learning sessions for various topics. It is stated that online Math help session is well-organized and hence, students can take this session and improve their performance in tests.

Given equation is quadratic equation, so let us solve by factorization method

2$x^2$ - 11 $x$ - 21 = 0

2$x^2$ - 14 $x$ + 3$x$ - 21 = 0

2$x$($x$ - 7) + 3($x$ - 7) = 0

(2$x$ + 3)($x$ - 7) = 0

By Zero Product Rule

2$x$ + 3 = o or $x$ - 7 = 0

$x$ = $\frac{-3}{2}$ or $x$ = 7

5x + 10m = 6 for x = 1

Put x = 1 in 5x + 10m = 6

5 $\times$ 1 + 10m = 6

5 + 10m = 6

10m = 1

m = $\frac{1}{10}$

The word 'OFFICES' consists of 7 letters out of which letter 'F' comes twice.

The total number of arrangements = $\frac{7!}{2!}$ = $\frac{7 . 6 . 5 . 4 . 3. 2!}{2!}$ = 2520

The letters of the word 'OFFICES' can be arranged in 2520 ways.