- 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, \(\triangle=\bigcirc\)

- 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 \(5\).

- 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 \(b=12\)

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

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

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

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

- 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.

A \(5\)

B \(7\)

C \(4\)

D \(3\)

- 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.

- 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:**

- \(x+5=20\) in equivalent to \(x=15\)
- \(3x-4=14\) is equivalent to \(3x=18\)
- \(2+(x+4)=9\) 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: \(2x(1+4)=20\)

A \(4x-4=14\)

B \(4x=14\)

C \(4x=16\)

D \(2x-2=14\)

- The substitution
- 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:

\(x=3\)

\(y=-2\)

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

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

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

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

A \(60\)

B \(61\)

C \(72\)

D \(59\)