Question

(vi) 3 (y + 3) + 4 = 28
y - 2​

Answer 1

Question

• 3(y + 3) + 4 = 28y - 2

Solution

$\sf\pink{⟶}$ In this question, we have to find the value of y.

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$\tt:\implies\: \: \: \: \: \: \: \: {3y + 9 + 4 = 28y - 2}$

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$\tt:\implies\: \: \: \: \: \: \: \: {3y + 13 = 28y - 2}$

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$\tt:\implies\: \: \: \: \: \: \: \: {28y - 3y = 13 + 2}$

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$\tt:\implies\: \: \: \: \: \: \: \: {25y = 15}$

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$\tt:\implies\: \: \: \: \: \: \: \: {y = \cancel{\dfrac{15}{25}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {y = \dfrac{3}{5}}$

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• Value of y is $\bold{\dfrac{3}{5}}$

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Answer 2

Step-by-step explanation:

$1.distribute \\ \\ 3(y + 3) + = 28y - 2 \\ 3y + 9 + 4 = 28y - 2 \\ \\ 2.add \: the \: numbers \\ \\ 3y + 9 + = 28y - 2 \\ 3y + 13 = 28y - 2 \\ \\ 3.simplify \\ \\ subtract \: the \: numbers \\ \\ 3y + 13 - 13 = 28y - 2 - 13 \\ 3y = 28y - 2 - 13 \\ \\ subtract \: the \: numbers \\ \\ 3y = 28y - 2 - 13 \\ 3y = 28y - 15 \\ 3y = 28y - 15 \\ \\ subtract \: 28y \: from \: \: both \: sides \: of \: the \: equation \: \\ \\ 3y = 28y - 15 \\ 3y - 28y = 28y - 15 - 28y \\ \\ simplify \\ \\ combine \: like \: terms \: \\ \\ 3y - 28y = 28y - 15 - 28y \\ - 25y = 28y - 15 - 28y \\ \\ combine \: like \: terms \: \\ \\ - 25y = 28y - 15 - 28y \\ - 25y = - 15 \\ - 25y = - 15 \\ \\ divide \: both \: sides \: of \: the \: equation \: by \: the \: same \: term \: \\ \\ - 25y = - 15 \\ \frac{ - 25y}{ - 25} = \frac{ - 15}{ - 25} \\ \\ simplify \\ \\ cancel \: terms \: that \: are \: in \: both \: the \: numerator \: and \: denominator \\ \\ \frac{ - 25y}{ - 25 } = \frac{ - 15}{ - 25} \\ y = \frac{ - 15}{ - 25} \\ \\ divide \: the \: numbers \: \\ \\ y = \frac{ - 15}{ - 25} \\ y = \frac{3}{5} \\ y = \frac{3}{5} \: solution \: is \: a \: = y = \frac{3}{5}$