Solve 7/3+sqrt{2}+7/3-sqrt{2}=6

$$\frac{ 7 }{ 3+ \sqrt{ 2 } } + \frac{ 7 }{ 3- \sqrt{ 2 } } =6$$

Verify

Solution Steps

Rationalize the denominator of \frac{7}{3+\sqrt{2}} by multiplying numerator and denominator by 3-\sqrt{2}.

$$\frac{7\left(3-\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+\frac{7}{3-\sqrt{2}}=6$$

Consider \left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{7\left(3-\sqrt{2}\right)}{3^{2}-\left(\sqrt{2}\right)^{2}}+\frac{7}{3-\sqrt{2}}=6$$

Square 3. Square \sqrt{2}.

$$\frac{7\left(3-\sqrt{2}\right)}{9-2}+\frac{7}{3-\sqrt{2}}=6$$

Subtract 2 from 9 to get 7.

$$\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{7}{3-\sqrt{2}}=6$$

Cancel out 7 and 7.

$$3-\sqrt{2}+\frac{7}{3-\sqrt{2}}=6$$

Rationalize the denominator of \frac{7}{3-\sqrt{2}} by multiplying numerator and denominator by 3+\sqrt{2}.

$$3-\sqrt{2}+\frac{7\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}=6$$

Consider \left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$3-\sqrt{2}+\frac{7\left(3+\sqrt{2}\right)}{3^{2}-\left(\sqrt{2}\right)^{2}}=6$$

Square 3. Square \sqrt{2}.

$$3-\sqrt{2}+\frac{7\left(3+\sqrt{2}\right)}{9-2}=6$$

Subtract 2 from 9 to get 7.

$$3-\sqrt{2}+\frac{7\left(3+\sqrt{2}\right)}{7}=6$$

Cancel out 7 and 7.

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

Add 3 and 3 to get 6.

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

Combine -\sqrt{2} and \sqrt{2} to get 0.

$$6=6$$

Compare 6 and 6.

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