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engelkil000 engelkil000

22-12-2021
Mathematics

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jimrgrant1 jimrgrant1 · Answer 1 · 2021-12-22T15:26:13+01:00

jimrgrant1 jimrgrant1

22-12-2021

Answer:

b = [tex]\frac{7}{12}[/tex]

Step-by-step explanation:

b + [tex]\frac{2}{3}[/tex] = 1 [tex]\frac{1}{4}[/tex] ← change to an improper fraction

b + [tex]\frac{2}{3}[/tex] = [tex]\frac{5}{4}[/tex]

Multiply through by 12 ( the LCM of 3 and 4 ) to clear the fractions

12b + 8 = 15 ( subtract 8 from both sides )

12b = 7 ( divide both sides by 12 )

b = [tex]\frac{7}{12}[/tex]

Answer Link

AadhPrit AadhPrit · Answer 2 · 2021-12-27T03:51:22+01:00

Answer:

The value of b is 7/12.

Step-by-step explanation:

Question :

Solve for b.

[tex]{\implies{\sf{b + \dfrac{2}{3} = 1 \dfrac{1}{4}}}}[/tex]

Enter your answer as a fraction in simplest form in the box.

[tex]\implies{\sf{b = \square}}[/tex]

[tex]\begin{gathered}\end{gathered}[/tex]

Solution :

[tex]{\implies{\sf{b + \dfrac{2}{3} = 1 \dfrac{1}{4}}}}[/tex]

Converting the mixed fractions into improper fraction.

[tex]{\implies{\sf{b + \dfrac{2}{3} = \dfrac{(1 \times 4) + 1}{4}}}}[/tex]

[tex]{\implies{\sf{b + \dfrac{2}{3} = \dfrac{(4)+ 1}{4}}}}[/tex]

[tex]{\implies{\sf{b + \dfrac{2}{3} = \dfrac{4+ 1}{4}}}}[/tex]

[tex]{\implies{\sf{b + \dfrac{2}{3} = \dfrac{5}{4}}}}[/tex]

Now, transporting LHS to RHS.

[tex]{\implies{\sf{b= \dfrac{5}{4} - \dfrac{2}{3}}}}[/tex]

Taking LCM of denominators and subtracting.

[tex]{\implies{\sf{b= \dfrac{(5 \times 3) - (2 \times 4)}{12}}}}[/tex]

[tex]{\implies{\sf{b= \dfrac{(15) - (8)}{12}}}}[/tex]

[tex]{\implies{\sf{b= \dfrac{15 - 8}{12}}}}[/tex]

[tex]{\implies{\sf{b= \dfrac{7}{12}}}}[/tex]

[tex]{\star{\red{\underline{\boxed{\sf{b= \dfrac{7}{12}}}}}}}[/tex]

Hence, the value of b is 7/12.

[tex]\begin{gathered}\end{gathered}[/tex]

Verification :

[tex]{\implies{\sf{b + \dfrac{2}{3} = 1 \dfrac{1}{4}}}}[/tex]

Substituting the value of (b=7/12)

[tex]{\implies{\sf{ \dfrac{7}{12} + \dfrac{2}{3} = 1 \dfrac{1}{4}}}}[/tex]

Converting mixed fractions into improper fraction

[tex]{\implies{\sf{ \dfrac{7}{12} + \dfrac{2}{3} = \dfrac{(1 \times 4) + 1}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{7}{12} + \dfrac{2}{3} = \dfrac{(4) + 1}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{7}{12} + \dfrac{2}{3} = \dfrac{4 + 1}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{7}{12} + \dfrac{2}{3} = \dfrac{5}{4}}}}[/tex]

Taking LCM of denominators in LHS and adding.

[tex]{\implies{\sf{ \dfrac{(7 \times 1) + (2 \times 4)}{12} = \dfrac{5}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{(7) + (8)}{12} = \dfrac{5}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{7 + 8}{12} = \dfrac{5}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{15}{12} = \dfrac{5}{4}}}}[/tex]

Cutting the fraction to simplest form.

[tex]{\implies{\sf{ \cancel{\dfrac{15}{12}} = \dfrac{5}{4}}}}[/tex]

[tex]{\implies{\sf{ \dfrac{5}{4}= \dfrac{5}{4}}}}[/tex]

[tex]{\star{\red{\underline{\boxed{\sf{LHS = RHS}}}}}}[/tex]

Hence Verified!

[tex]\underline{\rule{220pt}{3.5pt}}[/tex]