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Students who find it difficult to keep up with math lessons in school can enlist the help of live tutors who are experienced and know the subject well. Students work with tutors daily or weekly, as per their convenience and receive real-time help. These virtual sessions allow students to clear all their doubts and get explanations on topics that are not clear. Tutors also provide 6th grade math problems and answers enabling students to learn on the spot. Math word problems for 6th grade are also taught and students can ask their tutor when they have trouble with any words or phrases.### Solved Examples

**Question 1:**Solve (2x + y - 3z)

^{2}

**Solution:**

Given, (2x + y - 3z)

(a + b + c)

=> (2x + y - 3z)

= 4x

=> (2x + y - 3z)

^{2}We know(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca=> (2x + y - 3z)

^{2 }= (2x)^{2}+ (y)^{2}+ (-3z)^{2}+ 2 * 2x * y + 2 * y * (- 3z) + 2 * (- 3z) * 2x= 4x

^{2}+ y^{2}+ 9z^{ 2}+ 4xy - 6yz - 6zx=> (2x + y - 3z)

^{2}= 4x^{2}+ y^{2}+ 9z^{ 2}+ 4xy - 6yz - 6zx**Question 2:**Evaluate the expression when x = 1.

12x + 3y = 16

**Solution:**

Given, 12x + 3y = 16

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

**Question 3:**Find the Lowest common multiple of 36 and 24 having GCD 12.

**Solution:**

Two numbers are 36 and 24

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>

**LCM = 72**

### Solved Examples

**Question 1:**Find the perimeter of a rectangular garden with a length of 25 feet and width of 20 feet.

**Solution:**

Length of a rectangular garden = 25

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,

**Perimeter of rectangular garden is 90 feet****Question 2:**Find the two consecutive integers such that total of ten times the second integer and thrice the first integer is thirty-six.

**Solution:**

Let first integer = x

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

**Question 3:**The sum of two numbers is 10 and the difference is 4. What are the numbers?

**Solution:**

Let first number = x

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,

**First number = 7**

Second number = 3Second number = 3