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## Help Me Solve a Math Problem

## Solved Examples

**Question 1:**Verify that |x + y| $\leq$ |x| + |y| for x = $\frac{-4}{7}$, y = $\frac{-4}{3}$

**Solution:**

Since x = $\frac{-4}{7}$, y = $\frac{-4}{3}$

Step 1:

|x + y| = | $\frac{-4}{7}$ + $\frac{-4}{3}$|

= |$\frac{-12 - 28}{21}$|

= |$\frac{-40}{21}$|

= $\frac{40}{21}$

=> |x + y| = $\frac{40}{21}$

Step 2:

|x| + |y| = |$\frac{-4}{7}$| + |$\frac{-4}{3}$|

= $\frac{4}{7}$ + $\frac{4}{3}$

= $\frac{12 + 28}{21}$

= $\frac{40}{21}$

=> |x| + |y| = $\frac{40}{21}$

Therefore |x + y| = |x| + |y|, is true for x = $\frac{-4}{7}$, y = $\frac{-4}{3}$.

Step 1:

|x + y| = | $\frac{-4}{7}$ + $\frac{-4}{3}$|

= |$\frac{-12 - 28}{21}$|

= |$\frac{-40}{21}$|

= $\frac{40}{21}$

=> |x + y| = $\frac{40}{21}$

Step 2:

|x| + |y| = |$\frac{-4}{7}$| + |$\frac{-4}{3}$|

= $\frac{4}{7}$ + $\frac{4}{3}$

= $\frac{12 + 28}{21}$

= $\frac{40}{21}$

=> |x| + |y| = $\frac{40}{21}$

Therefore |x + y| = |x| + |y|, is true for x = $\frac{-4}{7}$, y = $\frac{-4}{3}$.

**Question 2:**Express $0.\bar{28}$ as a rational number.

**Solution:**

Given rational number is $0.\bar{28}$

Multiply $0.\bar{28}$ by 1

1 * $0.\bar{28}$ = 0.28282828............ ................................(1)

Multiply $0.\bar{28}$ by 100

100 * $0.\bar{28}$ = 28.2828282......... .................................(2)

(Decimal two places right)

Subtract (1) from (2)

100 * $0.\bar{28}$ - (1 * $0.\bar{28}$) = 28.2828282......... - 0.28282828............

or (100 - 1)$0.\bar{28}$ = 28

or 99 * $0.\bar{28}$ = 28

=> $0.\bar{28}$ = $\frac{28}{99}$.

=> $0.\bar{28}$ is a rational number.

Step 1:Step 1:

Multiply $0.\bar{28}$ by 1

1 * $0.\bar{28}$ = 0.28282828............ ................................(1)

Multiply $0.\bar{28}$ by 100

100 * $0.\bar{28}$ = 28.2828282......... .................................(2)

(Decimal two places right)

Step 2:Step 2:

Subtract (1) from (2)

100 * $0.\bar{28}$ - (1 * $0.\bar{28}$) = 28.2828282......... - 0.28282828............

or (100 - 1)$0.\bar{28}$ = 28

or 99 * $0.\bar{28}$ = 28

=> $0.\bar{28}$ = $\frac{28}{99}$.

=> $0.\bar{28}$ is a rational number.

**Question 3:**Solve $\frac{11}{7}$ = x + y and $\frac{11}{2}$ = -x + y

**Solution:**

Given equations

$\frac{11}{7}$ = x + y .....................(1)

$\frac{11}{2}$ = -x + y .......................(2)

Add (1) and (2)

=> x + y - x + y = $\frac{11}{7}$ + $\frac{11}{2}$

=> 2y = $\frac{22 + 77}{14}$

=> 2y = $\frac{99}{14}$

=> y = $\frac{99}{28}$

Put the value of y in (1)

=> $\frac{11}{7}$ = x + $\frac{99}{28}$

=> x = $\frac{11}{7}$ - $\frac{99}{28}$

=> x = $\frac{44 - 99}{28}$

=> x = $\frac{-55}{28}$

$\frac{11}{7}$ = x + y .....................(1)

$\frac{11}{2}$ = -x + y .......................(2)

Step 1:Step 1:

Add (1) and (2)

=> x + y - x + y = $\frac{11}{7}$ + $\frac{11}{2}$

=> 2y = $\frac{22 + 77}{14}$

=> 2y = $\frac{99}{14}$

=> y = $\frac{99}{28}$

Step 2:Step 2:

Put the value of y in (1)

=> $\frac{11}{7}$ = x + $\frac{99}{28}$

=> x = $\frac{11}{7}$ - $\frac{99}{28}$

=> x = $\frac{44 - 99}{28}$

=> x = $\frac{-55}{28}$

**Answer:**x = $\frac{-55}{28}$ and y = $\frac{99}{28}$.