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Rus_ich [418]
2 years ago
8

Can i get some help :

Mathematics
2 answers:
Charra [1.4K]2 years ago
5 0

Answer:

b = \frac{7}{12}

Step-by-step explanation:

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

b + \frac{2}{3} = \frac{5}{4}

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 = \frac{7}{12}

dimulka [17.4K]2 years ago
5 0

Answer:

The value of b is <u>7/12</u>.

Step-by-step explanation:

<u>Question</u> :

Solve for b.

{\implies{\sf{b +  \dfrac{2}{3} = 1 \dfrac{1}{4}}}}

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

\implies{\sf{b =  \square}}

\begin{gathered}\end{gathered}

<u>Solution</u> :

{\implies{\sf{b +  \dfrac{2}{3} = 1 \dfrac{1}{4}}}}

Converting the mixed fractions into improper fraction.

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

{\implies{\sf{b +  \dfrac{2}{3} = \dfrac{(4)+ 1}{4}}}}

{\implies{\sf{b +  \dfrac{2}{3} = \dfrac{4+ 1}{4}}}}

{\implies{\sf{b +  \dfrac{2}{3} = \dfrac{5}{4}}}}

Now, transporting LHS to RHS.

{\implies{\sf{b= \dfrac{5}{4} -  \dfrac{2}{3}}}}

Taking LCM of denominators and subtracting.

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

{\implies{\sf{b= \dfrac{(15) - (8)}{12}}}}

{\implies{\sf{b= \dfrac{15 - 8}{12}}}}

{\implies{\sf{b= \dfrac{7}{12}}}}

{\star{\red{\underline{\boxed{\sf{b= \dfrac{7}{12}}}}}}}

Hence, the value of b is 7/12.

\begin{gathered}\end{gathered}

<u>Verification</u> :

{\implies{\sf{b +  \dfrac{2}{3} = 1 \dfrac{1}{4}}}}

Substituting the value of (b=7/12)

{\implies{\sf{ \dfrac{7}{12}  +  \dfrac{2}{3} = 1 \dfrac{1}{4}}}}

Converting mixed fractions into improper fraction

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

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

{\implies{\sf{ \dfrac{7}{12}  +  \dfrac{2}{3} = \dfrac{4 + 1}{4}}}}

{\implies{\sf{ \dfrac{7}{12}  +  \dfrac{2}{3} = \dfrac{5}{4}}}}

Taking LCM of denominators in LHS and adding.

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

{\implies{\sf{ \dfrac{(7) + (8)}{12} = \dfrac{5}{4}}}}

{\implies{\sf{ \dfrac{7 + 8}{12} = \dfrac{5}{4}}}}

{\implies{\sf{ \dfrac{15}{12} = \dfrac{5}{4}}}}

Cutting the fraction to simplest form.

{\implies{\sf{ \cancel{\dfrac{15}{12}} = \dfrac{5}{4}}}}

{\implies{\sf{ \dfrac{5}{4}= \dfrac{5}{4}}}}

{\star{\red{\underline{\boxed{\sf{LHS = RHS}}}}}}

Hence Verified!

\underline{\rule{220pt}{3.5pt}}

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