Informative line

Solution Of Linear Equations In One Variable

Applications of Equations (Related to Quantity)

  • Equations can be used to solve various problems.
  • There are many cases where we can form equations to find the value of an unknown quantity, like in word problems.
  • Solve the real world and mathematical problems by writing and solving equations of the form:  \(x+p=q\) and \(px=q\) for the cases in which \(p,\;q\) and \(x\) are all non-negative rational numbers.

Real world problems related to quantity

To solve these types of word problems, follow the given steps:

(i) Write an equation from the given data.

(ii) Solve that equation.

Ex: Jacob has \(4\) marbles more than Marc. If they have \(30\) marbles altogether, find the number of marbles Marc has.

To solve this example:

(i) Write an equation.

Let, Marc has \(x\) marbles.

Jacob has \(4\) marbles more than Marc.

Here, the word, ''more than'' represents addition.

That means Jacob has "\(x+4\) " marbles.

The word, ''altogether''  also represents addition.

Hence, total marbles =  \(x+x+4\).

The result is \(30\).

\(\Rightarrow\;x+x+4=30\)

It is the required equation.

(ii) Now, solve the equation.

\(x+x+4=30\)

\(\Rightarrow\;2x+4=30\)

To find \(x,\;4\) should be removed by using the inverse operation of addition.

\(\therefore\) Subtracting \(4\) from both the sides

\(\Rightarrow\;2x+4-4=30-4\)

\(\Rightarrow\;2x=26\)

Now, \(2\) should also be removed to find \(x\) by using the inverse operation of multiplication.

\(\therefore\) Dividing by \(2\) on both the sides

\(\Rightarrow\;\dfrac{2x}{2}=\dfrac{26}{2}\)

\(\Rightarrow\;x=13\)

That means Marc has \(13\) marbles.

Illustration Questions

When Maria divided chocolates among her \(10\) friends, each got \(5\) chocolates. How many chocolates did she distribute?

A \(30\)

B \(50\)

C \(2\)

D \(15\)

×

Let the number of chocolates distributed be \(x.\)

Writing an equation from the given data.

Maria divided \(x\) chocolates among her \(10\) friends.

That means, \(\dfrac{x}{10}\)

Each friend got \(5\) chocolates, which means the result is \(5.\)

So we can write, \(\dfrac{x}{10}=5\)

Solving the equation to find \(x.\)

\(\dfrac{x}{10}=5\)

\(10\) should be removed to find \(x\) by using the inverse operation of division.

\(\therefore\) Multiplying by \(10\) on both the sides 

\(\Rightarrow\;\dfrac{x}{10}×10=5×10\)

\(\Rightarrow\;x=50\)

That means Maria distributed \(50\) chocolates.

Hence, option (B) is correct.

When Maria divided chocolates among her \(10\) friends, each got \(5\) chocolates. How many chocolates did she distribute?

A

\(30\)

.

B

\(50\)

C

\(2\)

D

\(15\)

Option B is Correct

Applications of Equations (Related to Cost)

  • Consider the given situation.

Sam bought \(8\) pens and paid \($30\) to the shopkeeper. If he got back \($6\) as balance, what is the price of one pen?

To solve this example:

(i) Write an equation.

The price of one pen is unknown, so let it be \(x.\) 

\(\therefore\) Price of \(8\) pens \(=8x\)

Sam paid \($30\) and got back \($6\), so the total amount he paid 

\(=$30-$6\)

\(=$24\)

Our result is \($24.\)

Now the statement is, the total cost of \(8\) pens is  \($24.\)

In the equation form, it can be written as:

 \(8x=24\)

(ii) Solve the equation.

\(8x=24\)

To find \(x,\) 8 should be removed by using the inverse operation of multiplication.

\(\therefore\) Dividing by \(8\) on both the sides

\(\Rightarrow\;\dfrac{8x}{8}=\dfrac{24}{8}\)

\(\Rightarrow\;x=3\)

That means the price of each pen is \($3\).

Illustration Questions

Carl bought pens costing \($3\) each and a book costing \($23\). If the total cost of all the items is \($50\), find the number of pens he bought.

A \(20\)

B \(30\)

C \(10\)

D \(9\)

×

Let the number of pens Carl bought be \(x.\)

Writing an equation.

Cost of one pen is \($3\).

So, the cost of \(x\) pens = \(3x\)

According to the statement,

Carl bought pens that cost \(3x\) and a book that costs \($23\).

That means, he bought items for  \(3x+23\)

\(\because\) The total cost is \($50\),

\(\therefore\) we can write, 

\(3x+23=50\)

This is the required equation.

Solving the equation to find \(x.\)

\(3x+23=50\)

\(23\) should be removed to find \(x\) by using the inverse operation of addition.

\(\therefore\) Subtracting \(23\) from both the sides

\(\Rightarrow\;3x+23-23=50-23\)

\(\Rightarrow\;3x=27\)

Now, \(3\) should also be removed by using the inverse operation of multiplication.

\(\therefore\) Dividing by \(3\) on both the sides

\(\dfrac{3x}{3}=\dfrac{27}{3}\)

\(\Rightarrow x=9\)

That means the number of pens Carl bought \(=9\)

Hence, option (D) is correct.

Carl bought pens costing \($3\) each and a book costing \($23\). If the total cost of all the items is \($50\), find the number of pens he bought.

A

\(20\)

.

B

\(30\)

C

\(10\)

D

\(9\)

Option D is Correct

Solving Equations using Balance Diagrams

  • An equation is like a weighing scale which shows that two things are equal.
  • We can use a balance diagram to represent simple linear equations which can be solved easily.
  • In a balanced situation, the two sides of the balance diagram should hold equal weight. 
  • To find out how much the object weighs, we can:

(i) Add the same amount to both the sides.

(ii) Take away the same amount from both the sides.

 

Consider the following example:

Here, we are provided with a scale.

\(\bigcirc\) (a ball) represents \(x\) and  \(\Box\) (a box) represents \(1.\)

Using the representation and the scale given, we have to find how many boxes balance a ball.

  • For this, we have to write expressions for both sides of the balancing scale.
  • For the left side, we can write "\(x+4\)" because it has \(1\) ball and \(4\) boxes.
  • For the right side, we can write "\(6\)" because it has \(6\) boxes.
  • Now, balance both the expressions by an equals sign.

\(x+4=6\)

  • To balance \(x\)  (a ball), \(4\) should be removed.
  • Subtract \(4\) from both the sides.

\(\Rightarrow\;x+4-4=6-4\)

\(\Rightarrow\;x=2\)

That means  \(2\) boxes balances a ball.

Now, the situation is represented as shown in the figure.

We can observe that the scale is balanced.

Illustration Questions

The scale below is balanced. Using the representation and the given scale, which is the correct option that could be placed on the right side of the following scale to balance a \(\Box\) ?

A

B

C

D

×

To balance a \(\Box\) (square), write expressions for both sides of the balancing scale.

\(\Box\) represents \(x\) and \(\triangle\) represents \(1.\)

For the left side, we can write "\(3x+4\)" because it has \(3\,\) squares \((\Box)\)  and \(4\,\) triangles \((\triangle)\).

For the right side, we can write "\(2x+7\)" because it has \(2\,\) squares \((\Box)\) and \(7\,\) triangles \((\triangle)\).

 Now, balance both the expressions by an equals sign.

\(3x+4=2x+7\)

To get the solution of the equation, the variable should be left alone on one side.

\(3x+4=2x+7\)

Subtract \(2x\) from both the sides

\(\Rightarrow\;3x+4-2x=2x+7-2x\)

\(\Rightarrow\;3x-2x+4=2x-2x+7\)

\(\Rightarrow\;x+4=7\)

To find \(x (\Box),\;4\) should be removed.

Subtract \(4\) from both sides of the equation.

\(\Rightarrow\;x+4-4=7-4\)

\(\Rightarrow\;x=3\)

That means a square  \((\Box)\)  balances  \(3\,\) triangles \((\triangle)\) .

The situation becomes,

image

Hence, option (A) is correct.

The scale below is balanced. Using the representation and the given scale, which is the correct option that could be placed on the right side of the following scale to balance a \(\Box\) ?

image
A image
B image
C image
D image

Option A is Correct

Illustration Questions

Carl bought two aquariums, one for the drawing room and another one for the bedroom. He paid \(15\%\) as shipping charges. The total amount paid by him is \($\; 345\). What is the actual cost of the aquariums?

A \($\;300\)

B \($\;45\)

C \($\;200\)

D \($\;35\)

×

Let the total actual cost of the aquariums be \($\;x\).

The total amount paid by him is \($\;345\).

Shipping fees is \(15\%\) of the actual cost =  \(\dfrac {15}{100}x\)

 

Thus, the equation becomes

\(x+\dfrac{15}{100}x = 345\)

\(x\left(1+\dfrac{15}{100}\right)= 345\)

\(x\left(\dfrac{100+15}{100}\right)= 345\)

\(x\left(\dfrac{115}{100}\right)= 345\)

\(x = \dfrac{345×100}{115}\)

\(x = 300\)

Thus, the total actual cost of the aquariums is \($\;300\).

Hence, option (A) is correct.

Carl bought two aquariums, one for the drawing room and another one for the bedroom. He paid \(15\%\) as shipping charges. The total amount paid by him is \($\; 345\). What is the actual cost of the aquariums?

A

\($\;300\)

.

B

\($\;45\)

C

\($\;200\)

D

\($\;35\)

Option A is Correct

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