Informative line

Solution Of Equations

Solving Models

  • To solve a given model, first we should convert it into an algebraic equation.
  • We can write algebraic equations through models.
  • Consider an example to understand it.

Write an algebraic equation for the following model.

\(\triangle\triangle\triangle\bigcirc\bigcirc=\bigcirc\bigcirc\bigcirc\bigcirc\triangle\)...............(1)

  • To write an algebraic equation, follow the given steps:

 

Step 1: Allot different variables to different figures.

Let \(\triangle=x\;\;\;\bigcirc=y\)

Step 2: Write individual expressions for both the groups that are present on both the sides of the equals sign.

 

 

 

For      \(\triangle\triangle\triangle\bigcirc\bigcirc\)

It is the group of three triangles 'and' two circles.

Here, 'and' represents addition (keyword).

\(\therefore\) Expression for this:

\(3x+2y\)     \((\triangle=x,\;\;\;\bigcirc=y)\)

 

Similarly,         \(\bigcirc\bigcirc\bigcirc\bigcirc\triangle\)

It is the group four circles 'and' one triangle

Here, 'and' represents addition (keyword).

\(\therefore\) Expression for this:

\(4y+x\)        \((\triangle=x,\;\;\;\bigcirc=y)\)

 

Step 3: Now put the expressions of both the groups in place of model to get an algebraic equation.

Thus, algebraic equation for the given model is:

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

  • Now solve the algebraic equation for \(x\) and  \(y.\)

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

 

Step 1: Since, \(x\) being added to \(4y\) therefore we will use inverse operation and subtract \(x\) from both the sides.

\(3x+2y-x=4y+x-x\)

\(2x+2y=4y\)

Step 2: Since, \(2y\) being added on both sides therefore we will use inverse operation and subtract \(2y\) from both the sides.

\(2x+2y-2y=4y-2y\)

\(2x=2y\)

or \(\dfrac{2x}{2}=\dfrac{2y}{2}\)

\(x=y\)

Step 3: Replace the values of \(x\) and \(y\).

Thus, \(\triangle=\bigcirc\)

Illustration Questions

An equation is modeled as shown below: Which one of the following options is correct?

A

B

C

D

×

Given model equation:

image

To solve a given model, first we should convert it into an algebraic equation.

Allot different variables to different figures.

Let

\(\bigcirc=x\) and \(\Box=y\)

Write individual expressions for both the groups that are present on both the sides of the equals sign.

For

image

It is a group of three circles 'and' three square.

Here, 'and' represents addition (keyword).

\(\therefore\) Expression for this:

\(3x+3y\)        \((\bigcirc=x,\;\;\;\Box=y)\)

 

For

image

It is a group of two squares 'and' five circles.

Here, 'and' represents addition (keyword).

\(\therefore\) Expression for this:

\(2y+5x\)        \((\Box=y,\;\;\;\bigcirc=x)\)

 

Now put the expressions of both the groups in the given model to get an algebraic equation.

Thus, algebraic equation for the given model is:

\(3x+3y=2y+5x\)

Solving this algebraic equation for \(x\) and \(y.\)

Since, \(3x\) is being added to \(3y\) therefore we will use inverse operation and subtract \(3x\) from both the sides.

\(3x+3y-3x=2y+5x-3x\)

\(3y=2y+2x\)

Since, \(2y\) is being added to \(2x\) therefore, we will use inverse operation and subtract \(2y\) from both the sides.

\(3y-2y=2y+2x-2y\)

\(y=2x\)

 

Replace the values of \(x\) and \(y.\)

Thus,  \(\Box=2\bigcirc\)

or

image

Hence, option (A) is correct.

An equation is modeled as shown below: Which one of the following options is correct?

image
A image
B image
C image
D image

Option A is Correct

Solving an Equation

  • Solving an equation means to find the value of a variable.
  • To solve an equation, we use the inverse operations to get only the variable on one side of the equation.
  • An inverse operation means opposite of an operation.
  • Inverse operation of addition is subtraction and vice versa.
  • Inverse operation of multiplication is division and vice versa.

For example:

Solve:  \(x+7=12\)

Here,

  • We need to find the value of the variable \((x)\).
  • Here, a variable \((x)\) is added to \(7\).
  • We will use inverse operation of addition, i.e. subtraction.

Subtract \(7\) from both the sides to get only \(x\) on the left-hand side.

\(x+7-7=12-7\)

\(\Rightarrow\;x+0=5\)

\(\Rightarrow\;x=5\)

  • So, answer is \(5\) or the value of \(x\) is \(5\).

Illustration Questions

Solve: \(3a+5=14\)

A \(3\)

B \(14\)

C \(5\)

D \(1\)

×

Given: \(3a+5=14\)

Here, \(5\) is added to \(3a\), so we will use inverse operation of addition, i.e. subtraction.

Thus, we subtract \(5\) from both the sides of the equation.

\(3a+5-5=14-5\)

\(3a+0=9\)

\(3a=9\)

To get only \('a'\) on one side of the equation, we use inverse operation again. 

Here, \(3\) is multiplied by \('a'\), it means the inverse operation of multiplication, i.e. division should be used.

Thus, we divide by \(3\) on both the sides of the equation.

\(\dfrac{3a}{3}=\dfrac{9}{3}\)

\(a=3\)

Thus, answer is \(3\) or the value of a is \(3.\)

Hence, option (A) is correct.

Solve: \(3a+5=14\)

A

\(3\)

.

B

\(14\)

C

\(5\)

D

\(1\)

Option A is Correct

Comparison of Both Sides of an Equation

  • We can determine the values of unknown constants by comparing both sides of the equation.
  • In the equation, coefficients of same variable on both the sides are equal.
  • We can determine the values of unknown constants by following the given steps:

Step-1 Identify the constants and variables.

Step-2 Compare the constant terms on both the sides of the equation.

Step-3 Compare coefficients of same variable on both the sides.

Example: \(ax+12=5x+b\)

If \(a\) and \(b\) are integers, determine their values.

  • First, identify the constants and variables.

Constants : \(a,\;12,\;5,\;b\)

Variable : \(x\)

  • Compare the constant terms on both the sides of the equation.

\(ax+12=5x+b\)

\(12=b\) ... (1)

  • Compare the coefficients of \(x\) on both the sides

\(ax+12=5x+b\)

\(a=5\) ... (2)

  • From (1) and (2), we get:

\(a=5\) and \(b=12\)

Illustration Questions

\((A+1)x+(B-5)y=9x+y+C\) Find the values of constants \(A,\;B\) and \(C.\)

A \(A=1,\;B=5,\;C=0\)

B \(A=9,\;B=1,\;C=0\)

C \(A=1,\;B=5,\;C=1\)

D \(A=8,\;B=6,\;C=0\)

×

Given : \((A+1)x+(B-5)y=9x+y+C\)

First, identify the constants and variables.

Constants : \((A+1),\;(B-5),\;9,\;C\)

Variables : \(x\) and \(y\)

Compare the constant terms on both the sides.

\((A+1)x+(B-5)y=9x+y+C\)

\(0=C\)  ... (1)

Compare the coefficients of \(x\) on both the sides.

\((A+1)x+(B-5)y=9x+y+C\)

\(A+1=9\)

\(\Rightarrow\;A=8\)  ... (2)

Compare the coefficients of \(y\) on both the sides.

\((A+1)x+(B-5)y=9x+y+C\)

\(B-5=1\)

\(\Rightarrow\;B=6\)  ... (3)

From (1), (2) and (3), we get :

\(A=8,\;B=6,\;C=0\)

Hence, option (D) is correct.

\((A+1)x+(B-5)y=9x+y+C\) Find the values of constants \(A,\;B\) and \(C.\)

A

\(A=1,\;B=5,\;C=0\)

.

B

\(A=9,\;B=1,\;C=0\)

C

\(A=1,\;B=5,\;C=1\)

D

\(A=8,\;B=6,\;C=0\)

Option D is Correct

Evaluation of an Equation

  • An equation in two variables can be evaluated if the value of one of the variables is known to us.

Steps to evaluate an equation:

  • Put the value of the known variable in the given equation.
  • Solve the equation for the other unknown variable.

For example:

In the equation, \(x+y=10\);

what is the value of \(x\)  if  \(y=5\) ?

  • Put the value of the known variable, \(y=5\),  in the equation,

\(x+y=10\)

\(\Rightarrow\;x+5=10\)

  • Solve the equation to find \(x.\)

\(x+5=10\)

\(\Rightarrow x+5-5=10-5\)

\(\Rightarrow x=5\)

It is the required solution.

Illustration Questions

If the equation is \(2x+3y=15\), what is the value of \(x\)  if  \(y=3\)?

A \(5\)

B \(7\)

C \(4\)

D \(3\)

×

To evaluate the equation,

\(2x+3y=15\)

Put the value of \(y=3\) in the equation

\(2x+3y=15\)

\(\Rightarrow\;2x+3(3)=15\)

\(\Rightarrow\;2x+9=15\)

Solve the equation to find \(x.\)

\(2x+9=15\)

\(\Rightarrow\;2x+9-9=15-9\)

\(\Rightarrow\;2x=6\)

\(\Rightarrow\;\dfrac{2x}{2}=\dfrac{6}{2}\)

\(\Rightarrow\;x=3\)

This is the required solution.

Hence, option (D) is correct.

If the equation is \(2x+3y=15\), what is the value of \(x\)  if  \(y=3\)?

A

\(5\)

.

B

\(7\)

C

\(4\)

D

\(3\)

Option D is Correct

Word Problems

  • Word Problems can be solved by evaluating equations, where value of one variable is given and we have to find the value of other variable.
  • We will explain the above statement with the help of an example.

For example: At present, age of father is \(18\) more than twice the age of son. If father's age is \(38\) years, calculate the age of his son at present. In the given situation, two variables are used:

Let \(x,\) represents the age of son.

\(y,\) represents the age of father.

Since, the age of the father is \(18\) more than twice the age of the son.

\(\therefore\;\text{Age of father}=2(\text{age of son})+18\)

\(\Rightarrow\;y=2x+18\)

To find the age of the son when his father's age is \(38\) years.

Put \(y=38\) in the equation

\(38=2(x)+18\)

\(2x=38-18\)

\(2x=20\)

\(\Rightarrow \;x=10\) years

This means, when the age of the father is \(38\) years old, his son is \(10\) years old.

 

Or,

Age of the father is \(18\) more than twice the age of the son.

Let the age of son is \(x.\) 

 

\(38-18=20\)

\(x+x=20\)

\(2x=20\)

\(x=10\)

Hence, son's age is \(10\) years.

Illustration Questions

Tim went to a store to buy some T-shirt and trousers. The price of each T-shirt is \(4\) less than the half the price of a trouser. Calculate the price of a T-shirt when the price of the trouser is \($50\).

A \($21\)

B \($25\)

C \($22\)

D \($15\)

×

Given statement:

Price of a T-shirt is \(4\) less than the half the price of a trouser.

Let \(a,\) represents the price of each trouser.

\(b,\) represents the price of each T-shirts.

According to the statement:

\(\text{Price of T-shirt}=\dfrac{\text{Price of trouser}}{2}-4\)

\(\Rightarrow\;b=\dfrac{a}{2}-4\)

To find the price of a T-shirt when price of the trouser is \($50\),

We put \(a=50\) in the equation:

\(\Rightarrow\;b=\dfrac{50}{2}-4\)

\(\Rightarrow\;b=25-4\)

\(\Rightarrow\;b=$\,21\)

This means, the price of T-shirt is \($21\).

Hence, option (A) is correct.

Tim went to a store to buy some T-shirt and trousers. The price of each T-shirt is \(4\) less than the half the price of a trouser. Calculate the price of a T-shirt when the price of the trouser is \($50\).

A

\($21\)

.

B

\($25\)

C

\($22\)

D

\($15\)

Option A is Correct

Equivalent Equation

  • Two equations are equivalent if both give same results.
  • When we solve an equation, every step of the solution is equivalent to each other.

For example: Solve the equation,

\(5x+8=18\)

\(5x+8=18\) is equivalent to \(5x+8-8=18-8\)

\(5x+8=18\) is equivalent to \(5x=10\)

\(5x+8=18\) is equivalent to \(x=2\)

Examples:

  1. \(x+5=20\) in equivalent to \(x=15\)
  2. \(3x-4=14\) is equivalent to \(3x=18\)
  3. \(2+(x+4)=9\) is equivalent to \((2+x)+4=9\)
  • We can also find equivalent equations by using properties.
  • For example, \(2x(4+1)=20\)

So, by using commutative property, the equivalent equation is: \(2x(1+4)=20\)

Illustration Questions

Which one of the following equations is equivalent to the equation:   \(2(2x-1)=14\)

A \(4x-4=14\)

B \(4x=14\)

C \(4x=16\)

D \(2x-2=14\)

×

Given: \(2(2x-1)=14\)

We can find the equivalent equation of the given equation by solving it.

\(2(2x-1)=14\)

\(4x-2=14\)

\(4x=16\)

 

 

We can see that option (C) is equivalent to the given equation.

Therefore \(2(2x-1)=14\) is equivalent to \(4x=16\)

 

 

Hence, option (C) is correct.

Which one of the following equations is equivalent to the equation:   \(2(2x-1)=14\)

A

\(4x-4=14\)

.

B

\(4x=14\)

C

\(4x=16\)

D

\(2x-2=14\)

Option C is Correct

Substitution Method

  • The substitution method is used to determine the value of a variable.
  • For substitution method, the number of variables used in the equations should be equal to the number of equations.

For substitution method, we follow the given steps -

Step-1 Substitute the value of the given variable in the equation and determine the value of another variable.

Step-2 Again substitute the value of the variable (which was found out in step \(1\)) in next equation.

Step-3 Repeat the process to determine the value of each variable.

Example:

\(3x+2y=5,\;x+z=4,\;z=1\)

Determine the values of \(x\) and \(y.\)

\(3x+2y=5\)  ... (1)

\(x+z=4\)  ... (2)

  • Substitute \(z=1\) in equation (2) and determine the value of \(x.\)

\(x+1=4\)

\(x=4-1\)

\(x=3\)  ... (3)

  • Substitute \(x=3\) in equation (1) and determine the value of \(y.\)

\(3(3)+2y=5\)

\(9+2y=5\)

\(2y=5-9\)

\(2y=-4\)

\(y=\dfrac{-4}{2}=-2\)

\(y=-2\)  ... (4)

  • From equations (3) and (4), we get:

\(x=3\)

\(y=-2\)

Illustration Questions

\(4x+3y=16,\;x+z=3,\;z=2\)   Determine the values of \(x\) and \(y.\)

A \(x=-1,\;y=4\)

B \(x=1,\;y=4\)

C \(x=1,\;y=-4\)

D \(x=-1,\;y=-4\)

×

Given: 

\(4x+3y=16\) ... (1)

\(x+z=3\) ... (2)

\(z=2\)

Substitute \(z=2\) in equation (2)

\(x+2=3\)

\(\Rightarrow\;x=1\) ... (3)

Substitute  \(x=1\) in equation (1)

\(4(1)+3y=16\)

\(\Rightarrow\;4+3y=16\)

\(\Rightarrow\;3y=16-4\)

\(\Rightarrow\;3y=12\)

\(\Rightarrow\;y=4\) ... (4)

From equations (3) and (4), we get:

\(x=1\)

\(y=4\)

Hence, option (B) is correct.

\(4x+3y=16,\;x+z=3,\;z=2\)   Determine the values of \(x\) and \(y.\)

A

\(x=-1,\;y=4\)

.

B

\(x=1,\;y=4\)

C

\(x=1,\;y=-4\)

D

\(x=-1,\;y=-4\)

Option B is Correct

Illustration Questions

What is the value of \(y\) in the given polynomial equation, if \(x = 3\) ? \(y = x^3 + 6x+16\)

A \(60\)

B \(61\)

C \(72\)

D \(59\)

×

Given :

\(y =x^3 + 6x+16\)

 \(x=3\)

We will put \(3\) in place of \(x\), to find the value of \(y\).

 

\(y=x^3+6x+16\)

\(y=(3)^3+6(3)+16\)

\(y = 3 ×3×3+6×3+16\)

\(y = 27+18+16\)

\(y = 61\)

Hence, option (B) is correct.

What is the value of \(y\) in the given polynomial equation, if \(x = 3\) ? \(y = x^3 + 6x+16\)

A

\(60\)

.

B

\(61\)

C

\(72\)

D

\(59\)

Option B is Correct

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