Excercise 4.3 Simple-Equations - NCERT Maths Class 7


Chapter 4 Ex.4.3 Question 1

Solve the following equations.

(a) \(\begin{align} 2y + \fracabc{2} = \frac{{37}}{2}\end{align} \)

(b)  \(\begin{align} 5t + 28 = 10\end{align} \)

(c) \(\begin{align} \frac{a}{5} + 3 = 2\end{align} \)

(d) \(\begin{align} \frac{q}{4} + 7 = 5\end{align} \)

(e) \(\begin{align} \frac{5}{2}x = - 5\end{align} \)

(f) \(\begin{align} \frac{5}{2}x = \frac{{25}}{4}\end{align} \)

(g) \(\begin{align} 7m + \frac{{19}}{2} = 13\end{align} \)

(h)  \(\begin{align} 6z + 10 = - 2\end{align} \)

(i) \(\begin{align} \frac{{3l}}{2} = \frac{2}{3}\end{align} \)

(j) \(\begin{align} \frac{{2b}}{3} - 5 = 3\end{align} \)

 

Solution

Video Solution

 

What is Known?

Equations.

What is unknown?

Solution of the equations (The value of the variable).

Reasoning:

To solve these equations, first transpose the variables on the one side and constants on the other side, then simplify them and get the value of variable.

Steps:

(a) \(\begin{align} 2y + \frac{5}{2} = \frac{{37}}{2}\end{align} \)

Transposing \(\begin{align} \frac{5}{2}\end{align} \) to R.H.S we get,

\[\begin{align}2y &= \frac{{37}}{2} - \frac{5}{2}\\2y &= \frac{{32}}{2} = 16\\y &= \frac{{16}}{2} = 8\end{align}\]

(b)  \(\begin{align} 5t + 28 = 10\end{align} \)

Transposing \(28\) to R.H.S we get,

\[\begin{align}5t &= 10 - 28\\5t &= - 18\\t &= \frac{{ - 18}}{5}\end{align}\]

(c) \(\begin{align} \frac{a}{5} + 3 = 2\end{align} \)

Transposing \(3\) to R.H.S we get,

\[\begin{align}\frac{a}{5} &= 2 - 3\\\frac{a}{5} &= - 1\\a &= - 5\end{align}\]

(d) \(\begin{align} \frac{q}{4} + 7 = 5\end{align} \)

Transposing \(7\) to R.H.S we get,

\[\begin{align}\frac{q}{4} &= 5 - 7\\\frac{q}{4} &= - 2\\q &= - 8\end{align}\]

(e) \(\begin{align} \frac{5}{2}x = - 5\end{align} \)

\[\begin{align}5x &= - 5 \times 2\\x &= \frac{{ - 10}}{5}\\x &= - 2\end{align}\]

(f) \(\begin{align} \frac{5}{2}x = \frac{{25}}{4}\end{align} \)

\[\begin{align}5x &= \frac{{25}}{4} \times 2\\x &= \frac{{25}}{{2 \times 5}}\\x &= \frac{5}{2}\end{align}\]

(g)  \(\begin{align} 7m + \frac{{19}}{2} = 13\end{align} \)

Transposing \(\begin{align} \frac{{19}}{2}\end{align} \) to the R.H.S.

\[\begin{align}7m&= 13 - \frac{{19}}{2}\\7m &= \frac{{26 - 19}}{2}\\7m &= \frac{7}{2}\\m &= \frac{7}{{2 \times 7}}\\m &= \frac{1}{2}\end{align}\]

(h)  \(\begin{align} 6z + 10 = - 2\end{align} \)

Transposing \(10\) to the R.H.S.

\[\begin{align}6z &= - 2 - 10\\z &= \frac{{ - 12}}{6}\\z& = - 2\end{align}\]

(i) \(\begin{align} \frac{{3l}}{2} = \frac{2}{3}\end{align} \)

\[\begin{align}l &= \frac{2}{3} \times \frac{2}{3}\\l &= \frac{4}{9}\end{align}\]

(j) \(\begin{align} \frac{{2b}}{3} - 5 = 3\end{align} \)

\[\begin{align}\frac{{2b}}{3} &= 3 + 5\\\frac{{2b}}{3} &= 8\\b &= 8 \times \frac{3}{2}\\b &= 12\end{align}\]

Chapter 4 Ex.4.3 Question 2

Solve the following equations.

(a) \(2(x + 4) = 12\)

(b) \(3(n\, – 5) = 21\)

(c) \(\begin{align}3\left( {n-5} \right) = - 21\end{align} \)

(d) \(\begin{align}- 4\left( {2 + x} \right) = 8\end{align} \)

(e) \(4(2 \,– x) = 8\)

 

Solution

Video Solution

 

What is Known?

Equations.

What is unknown?

The value of the variable.

Reasoning:

To solve these equations, transpose the variables on the one side and constants on the other side, and simplify them and get the value of variable.

Steps:

(a) \(2(x + 4) = 12\)

\[\begin{align} 2(x + 4) &= 12\\2x + 8 &= 12\\2x &= 12 – 8\\2x &= 4\\x& = \frac{4}{2}\; \text{or} \;x = 2\end{align} \]

(b) \(3(n \,– 5) = 21\)

\[\begin{align}3(n – 5) &= 21\\3n – 15 &= 21\\3n &= 21 + 15\\3n &= 36 \\ n &= \frac{{36}}{3} \,\text{or}\, n = 12\end{align} \]

(c) \(\begin{align} 3\left( {n-{\rm{ }}5} \right) = - 21\end{align} \)

\[\begin{align} 3\left( {n-{\rm{ }}5} \right) &= - 21 \\3n \,– 15 &=\, –21\\3n &=\, –21 + 15\\3n& = \,–6\\ n &= \frac{{ - 6}}{3}\,\text{or}\, n = \,–2\end{align} \]

(d) \(\begin{align}- 4\left( {2 + x} \right) = 8\end{align} \)

\[\begin{align} - 4\left( {2 + x} \right) &= 8\\– 8 \,– 4x &= 8\\–4x &= 8 + 8\\–4x &= 16\\x = \frac{{ - 16}}{4}& = \,–\,4\end{align} \]

(e)  \(4(2 \,– x) = 8\)

\[\begin{align}4(2 – x) &= 8\\8 – 4x &= 8 \\– 4x &= 8\,– 8 = 0 \,\text{or} \,x = 0\end{align}\]

Chapter 4 Ex.4.3 Question 3

Solve the following equations.

(a) \(4 = 5(p \,– 2) \)

(b) \(– 4 = 5(p\, – 2)\)

(c) \(16 = 4 + 3(t + 2)\)

(d) \(4 + 5(p - 1) =34\)

(e) \(0 = 16 + 4(m \,– 6)\)

 

Solution

Video Solution

 

What is Known?

Equations.

What is unknown?

The value of the variable.

Reasoning:

Transpose the variables on the one side and constants on the other side, then simplify them and get the value of variable.

Steps:

(a) \(4 = 5(p\, – 2) \)

\[\begin{align}4 &= 5(p\, – 2)\\4& = 5p \,– 10\\5p &= 4 + 10\end{align}\]

Therefore, \(\begin{align} p = \frac{{14}}{5}\end{align} \)

(b) \(– 4 = 5(p \,– 2)\)

\[\begin{align}– 4&= 5(p \,– 2)\\– 4 &= 5p \,– 10\\– 4 + 10 &= 5p\\6&= 5p\end{align}\]

Therefore, \(\begin{align} p = \frac{6}{5}\end{align} \)

(c) \(16 = 4 + 3(t + 2)\)

\[\begin{align}16 &= 4 + 3(t + 2)\\16 &= 4 + 3t + 6\\16 \,– 10 &= 3t\\6 &=3t\\ t &= \frac{6}{3} = 2\end{align}\]

(d) \(4 + 5(p - 1) =34\)

\[\begin{align} 4 + 5(p - 1) &=34\\4 + 5p \,– 5 &= 34\\5p \,– 1 &= 34\\5p &= 35\\p &= \frac{{35}}{5}\\p &= 7\end{align} \]

(e) \(0 = 16 + 4(m \,– 6)\)

\[\begin{align}0 &= 16 + 4(m \,– 6)\\0 &= 16 + 4m \,– 24\\8 &= 4m\\ m &= \frac{8}{4}\\m &= 2\end{align} \]

Chapter 4 Ex.4.3 Question 4

(a) Construct \(3\) equations starting with \(x = 2\)

(b) Construct \(3\) equations starting with \(x=-2\)

 

Solution

Video Solution

 

What is Known?

Value of the variables; \(x = 2\) and \(x=-2\)

What is unknown?

\(3\) equations starting with \(x = 2\) and \(3\) equations starting with \(x = 2\)

Reasoning:

You can get the equation by adding, multiplying or subtracting the same value on both sides of the equation.

Steps:

(a) \(3\) equations starting with \(x = 2\)

(i) \(x = 2\)

Multiplying both sides by \(10\),

\(10x = 20\)

Adding \(2\) to both sides ,

\[\begin{align}10x + 2 &= 20 + 2\\10x + 2 &= 22\end{align}\]

(ii) \(x = 2\)

multiplying both sides by \(5\),

\(5x = 10\)

subtracting \(3\) to both sides,

\[\begin{align}5x - 3 &= 10 - 3\\5x - 3 &= 7\end{align}\]

(iii) \(x = 2\)

multiplying both sides by \(2\),

\(2x = 4\)

subtracting \(3\) to both sides,

\[\begin{align}2x - 3 &= 4 - 3\\2x - 3 &= 1\end{align}\]

(b) \(3\) equations starting with \(x = -2\)

(i) \(x=-2\)

Multiplying both sides by \(3\),

\(3x = - 6\)

(ii) \(x=-2\)

Multiplying both sides by \(3\),

\(3x = - 6\)

Adding \(7 \) to both sides we get.

\[\begin{align}3x + 7 &= - 6 + 7\\ 3x + 7 &= 1\end{align}\]

(iii) \(x=-2\)

Multiplying both sides by \(3\),

\(3x = - 6\)

Adding \(10\) to both sides, we get.

\[\begin{align}3x + 10 &= - 6 + 10\\ 3x + 10& = 4\end{align}\]

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