Solve |(-2)/(9)+(-3)/(6)| <= |(-2)/(9)|+|(-3)/(6) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$| \frac{ -2 }{ 9 } + \frac{ -3 }{ 6 } | \leq | \frac{ -2 }{ 9 } |+| \frac{ -3 }{ 6 }$$

Verify

$\text{true}$

Solution Steps

Fraction $\frac{-2}{9}$ can be rewritten as $-\frac{2}{9}$ by extracting the negative sign.

$$|-\frac{2}{9}+\frac{-3}{6}|\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

Reduce the fraction $\frac{-3}{6}$ to lowest terms by extracting and canceling out $3$.

$$|-\frac{2}{9}-\frac{1}{2}|\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

Least common multiple of $9$ and $2$ is $18$. Convert $-\frac{2}{9}$ and $\frac{1}{2}$ to fractions with denominator $18$.

$$|-\frac{4}{18}-\frac{9}{18}|\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

Since $-\frac{4}{18}$ and $\frac{9}{18}$ have the same denominator, subtract them by subtracting their numerators.

$$|\frac{-4-9}{18}|\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

Subtract $9$ from $-4$ to get $-13$.

$$|-\frac{13}{18}|\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $-\frac{13}{18}$ is $\frac{13}{18}$.

$$\frac{13}{18}\leq |\frac{-2}{9}|+|\frac{-3}{6}|$$

Fraction $\frac{-2}{9}$ can be rewritten as $-\frac{2}{9}$ by extracting the negative sign.

$$\frac{13}{18}\leq |-\frac{2}{9}|+|\frac{-3}{6}|$$

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $-\frac{2}{9}$ is $\frac{2}{9}$.

$$\frac{13}{18}\leq \frac{2}{9}+|\frac{-3}{6}|$$

Reduce the fraction $\frac{-3}{6}$ to lowest terms by extracting and canceling out $3$.

$$\frac{13}{18}\leq \frac{2}{9}+|-\frac{1}{2}|$$

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $-\frac{1}{2}$ is $\frac{1}{2}$.

$$\frac{13}{18}\leq \frac{2}{9}+\frac{1}{2}$$

Least common multiple of $9$ and $2$ is $18$. Convert $\frac{2}{9}$ and $\frac{1}{2}$ to fractions with denominator $18$.

$$\frac{13}{18}\leq \frac{4}{18}+\frac{9}{18}$$

Since $\frac{4}{18}$ and $\frac{9}{18}$ have the same denominator, add them by adding their numerators.

$$\frac{13}{18}\leq \frac{4+9}{18}$$

Add $4$ and $9$ to get $13$.

$$\frac{13}{18}\leq \frac{13}{18}$$

Compare $\frac{13}{18}$ and $\frac{13}{18}$.

$$\text{true}$$