6. Simplify the ff. (a) $$ \frac{x}{4}+\frac{3}{7} $$ b) $$ \frac{2 x}{4+x}+\frac{x}{3} $$ (3) $$ \frac{a+1}{b} imes \frac{b^{2}}{a^{2}-1} $$

Question

6. Simplify the ff. (a) $$ \frac{x}{4}+\frac{3}{7} $$ b) $$ \frac{2 x}{4+x}+\frac{x}{3} $$ (3) $$ \frac{a+1}{b} 	imes \frac{b^{2}}{a^{2}-1} $$

UpStudy AI · Accepted Answer

Calculate or simplify the expression $$ (x/4)+(3/7) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{x}{4}+\frac{3}{7}$$
- step1: Reduce fractions to a common denominator:
  $$\frac{x	imes 7}{4	imes 7}+\frac{3	imes 4}{7	imes 4}$$
- step2: Multiply the numbers:
  $$\frac{x	imes 7}{28}+\frac{3	imes 4}{7	imes 4}$$
- step3: Multiply the numbers:
  $$\frac{x	imes 7}{28}+\frac{3	imes 4}{28}$$
- step4: Transform the expression:
  $$\frac{x	imes 7+3	imes 4}{28}$$
- step5: Reorder the terms:
  $$\frac{7x+3	imes 4}{28}$$
- step6: Multiply the numbers:
  $$\frac{7x+12}{28}$$
Calculate or simplify the expression $$ (2*x/(4+x))+(x/3) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\left(\frac{2x}{\left(4+xight)}ight)+\frac{x}{3}$$
- step1: Remove the parentheses:
  $$\left(\frac{2x}{4+x}ight)+\frac{x}{3}$$
- step2: Remove the parentheses:
  $$\frac{2x}{4+x}+\frac{x}{3}$$
- step3: Reduce fractions to a common denominator:
  $$\frac{2x	imes 3}{\left(4+xight)	imes 3}+\frac{x\left(4+xight)}{3\left(4+xight)}$$
- step4: Reorder the terms:
  $$\frac{2x	imes 3}{3\left(4+xight)}+\frac{x\left(4+xight)}{3\left(4+xight)}$$
- step5: Transform the expression:
  $$\frac{2x	imes 3+x\left(4+xight)}{3\left(4+xight)}$$
- step6: Multiply the terms:
  $$\frac{6x+x\left(4+xight)}{3\left(4+xight)}$$
- step7: Multiply the terms:
  $$\frac{6x+4x+x^{2}}{3\left(4+xight)}$$
- step8: Add the terms:
  $$\frac{10x+x^{2}}{3\left(4+xight)}$$
- step9: Simplify:
  $$\frac{10x+x^{2}}{12+3x}$$
Calculate or simplify the expression $$ ((a+1)/b)*((b^2)/(a^2-1)) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\left(\frac{\left(a+1ight)}{b}ight)\left(\frac{b^{2}}{\left(a^{2}-1ight)}ight)$$
- step1: Remove the parentheses:
  $$\left(\frac{a+1}{b}ight)\left(\frac{b^{2}}{a^{2}-1}ight)$$
- step2: Remove the parentheses:
  $$\frac{a+1}{b}	imes \left(\frac{b^{2}}{a^{2}-1}ight)$$
- step3: Remove the parentheses:
  $$\frac{a+1}{b}	imes \frac{b^{2}}{a^{2}-1}$$
- step4: Rewrite the expression:
  $$\frac{a+1}{b}	imes \frac{b^{2}}{\left(a+1ight)\left(a-1ight)}$$
- step5: Reduce the fraction:
  $$1	imes \frac{b}{a-1}$$
- step6: Multiply the terms:
  $$\frac{b}{a-1}$$
The simplified expressions are:
a) $$ \frac{x}{4}+\frac{3}{7} = \frac{7x+12}{28} $$
b) $$ \frac{2x}{4+x}+\frac{x}{3} = \frac{10x+x^{2}}{12+3x} $$
c) $$ \frac{a+1}{b} 	imes \frac{b^{2}}{a^{2}-1} = \frac{b}{a-1} $$
a) $$ \frac{x}{4} + \frac{3}{7} = \frac{7x + 12}{28} $$ b) $$ \frac{2x}{4+x} + \frac{x}{3} = \frac{10x + x^{2}}{12 + 3x} $$ c) $$ \frac{a+1}{b} 	imes \frac{b^{2}}{a^{2}-1} = \frac{b}{a-1} $$

6. Simplify the ff. (a) \( \frac{x}{4}+\frac{3}{7} \) b) \( \frac{2 x}{4+x}+\frac{x}{3} \) (3) \( \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \)

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