SSC GD Algebra

• Introduction to SSC GD Algebra
• Simple Algebraic Expressions
  • What are Variables and Constants?
  • Adding and Subtracting Expressions
• Linear Equations (One Variable)
  • The Balance Rule
  • Solving for X
• Basic Algebraic Identities (Shortcuts)
  • Identity 1: The Plus Shortcut
  • Identity 2: The Minus Shortcut
  • Identity 3: The Difference Shortcut

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Introduction to SSC GD Algebra

Algebra is like a puzzle game where we use letters (like $x$ or $y$) to stand for numbers we do not know yet. Learning SSC GD Algebra helps you solve these puzzles quickly.

• In algebra, we learn how to find the missing value.
• This topic is very important for the SSC GD Constable 2026 exam.
• We will keep the rules very simple so you can master SSC GD Algebra easily.

Simple Algebraic Expressions

What are Variables and Constants?

An expression is a mix of numbers and letters.

• Variables: These are the letters ($x,y,a,b$). Their value can change. Think of $x$ as a box where you can put any number inside.
• Constants: These are the plain numbers (5, 10, 100). Their value never changes.
• Example: In the expression $3x+5$, $x$ is the variable, and 3 and 5 are constants.

Adding and Subtracting Expressions

We can only add or subtract things that are the same.

• Think of it like fruits: You can add 3 apples ($3a$) and 2 apples ($2a$) to get 5 apples ($5a$).
• You cannot add 3 apples ($3a$) and 2 bananas ($2b$). They must stay separate.
• This rule is key to solving problems in SSC GD Algebra.

4 Solved Examples

Example 1: Find the value of $5x+3x-2x$.

Solution:

1. First, let’s understand that all these terms have the same variable ($x$). They are all “apples.”
2. We can just add and subtract the numbers in front of $x$.
3. We start with $5+3$.
   • $5+3=8$
4. Now we take the result and subtract 2.
   • $8-2=6$
5. So, the answer is $6x$.

Example 2: Simplify the expression $4a+7b-2a+b$.

Solution:

1. First, let’s group the terms that are the same. Group the ‘a’ terms and the ‘b’ terms.
   • ‘a’ terms: $4a$ and $-2a$
   • ‘b’ terms: $7b$ and $b$ (Remember $b$ means $1b$).
2. Now, let’s solve the ‘a’ terms first.
   • $4a-2a=2a$
3. Next, solve the ‘b’ terms.
   • $7b+1b=8b$
4. We cannot add $2a$ and $8b$ because they are different variables.
5. So, the answer is $2a+8b$.

Example 3: If $x=4$, what is the value of the expression $2x+10$?

Solution:

1. First, let’s understand that we need to replace the letter $x$ with the number 4.
2. The expression $2x$ means 2 multiplied by $x$. So, we write $2\times 4$.
3. Now, put the numbers in the expression:
   • $2\times 4+10$
4. Calculate the multiplication first (BODMAS rule).
   • $8+10$
5. So, the answer is 18. This is a common type of question in SSC GD Algebra.

Example 4: Multiply the expression $3(x+5)$.

Solution:

1. First, let’s understand that the number outside the bracket (3) must multiply everything inside the bracket.
2. Multiply 3 by $x$:
   • $3\times x=3x$
3. Multiply 3 by 5:
   • $3\times 5=15$
4. Now, put the results together with the plus sign.
5. So, the answer is $3x+15$.

Linear Equations (One Variable)

The Balance Rule

A linear equation is like a balance scale. The equals sign ($=$) means that the left side must weigh exactly the same as the right side.

• Example: $x+2=7$. We need to find the number $x$ that makes the sides equal.
• The main goal in SSC GD Algebra equations is to get the variable ($x$) all alone on one side.

Solving for X

When you move a number across the equals sign, you must change its sign (the opposite operation).

4 Solved Examples

Example 1: Solve for $x$: $x+9=15$.

Solution:

1. First, we want to get $x$ alone. We need to move the $+9$ to the other side.
2. When $+9$ moves across the equals sign, it becomes $-9$.
   • $x=15-9$
3. Now, calculate the subtraction.
   • $15-9=6$
4. So, the answer is $x=6$.

Example 2: Solve for $y$: $4y=20$.

Solution:

1. First, remember that $4y$ means 4 multiplied by $y$.
2. To get $y$ alone, we must move the 4. Since it is multiplying on the left, it must divide on the right.
   • $y=\frac{20}{4}$
3. Now, calculate the division.
   • $20÷4=5$
4. So, the answer is $y=5$. This type of problem is very common in SSC GD Algebra papers.

Example 3: Solve for $x$: $3x-5=13$.

Solution:

1. First, we move the number that is being added or subtracted. Move $-5$ to the right side. It becomes $+5$.
   • $3x=13+5$
2. Calculate the right side.
   • $3x=18$
3. Now, $3$ is multiplying $x$. Move the 3 to the right side by dividing.
   • $x=\frac{18}{3}$
4. Calculate the division.
   • $x=6$
5. So, the answer is $x=6$.

Example 4: Solve for $x$: $\frac{x}{2}+1=7$.

Solution:

1. First, move the $+1$ to the right side. It becomes $-1$.
   • $\frac{x}{2}=7-1$
2. Calculate the right side.
   • $\frac{x}{2}=6$
3. Now, 2 is dividing $x$. To move 2, we must multiply it on the right side.
   • $x=6\times 2$
4. Calculate the multiplication.
   • $x=12$
5. So, the answer is $x=12$. Practice these steps for your SSC GD Algebra preparation.

Basic Algebraic Identities (Shortcuts)

Identities are special formulas that are always true, no matter what numbers $a$ and $b$ are. They are shortcuts to save time during the SSC GD Constable exam.

Identity 1: The Plus Shortcut

This identity is used when you multiply $(a+b)$ by itself.

• Formula: $(a+b{)}^2=a^2+2ab+b^2$
• Remember: Square the first term ($a^2$), square the second term ($b^2$), and add two times the product of both terms ($2ab$).

Identity 2: The Minus Shortcut

This identity is used when you multiply $(a-b)$ by itself.

• Formula: $(a-b{)}^2=a^2-2ab+b^2$
• The only difference from the plus shortcut is the middle term: it is minus $2ab$.

Identity 3: The Difference Shortcut

This identity is used when you multiply $(a+b)$ by $(a-b)$.

• Formula: $a^2-b^2=(a+b)(a-b)$
• This is the fastest shortcut! If you see two squares subtracted, you just add the terms once and subtract them once, and then multiply the results. This is very useful in SSC GD Algebra.

4 Solved Examples

Example 1: Expand $(x+3{)}^2$ using the identity.

Solution:

1. First, identify $a$ and $b$. Here, $a=x$ and $b=3$.
2. Use the formula $(a+b{)}^2=a^2+2ab+b^2$.
3. Substitute $a$ and $b$ into the formula:
   • $x^2+2(x)(3)+3^2$
4. Calculate each part:
   • $x^2$ stays $x^2$.
   • $2(x)(3)=6x$.
   • $3^2=9$.
5. So, the answer is $x^2+6x+9$.

Example 2: Expand $(2y-5{)}^2$ using the identity.

Solution:

1. First, identify $a$ and $b$. Here, $a=2y$ and $b=5$.
2. Use the formula $(a-b{)}^2=a^2-2ab+b^2$.
3. Substitute $a$ and $b$ into the formula:
   • $(2y{)}^2-2(2y)(5)+5^2$
4. Calculate each part carefully:
   • $(2y{)}^2=2^2\times y^2=4y^2$.
   • $-2(2y)(5)=-20y$.
   • $5^2=25$.
5. So, the answer is $4y^2-20y+25$. Mastering these identities is key for SSC GD Algebra.

Example 3: Calculate ${101}^2$ using an algebraic identity.

Solution:

1. First, we can write 101 as a sum of easy numbers: $101=100+1$.
2. So, we need to calculate $(100+1{)}^2$. Here, $a=100$ and $b=1$.
3. Use the formula $(a+b{)}^2=a^2+2ab+b^2$.
4. Substitute the numbers:
   • $(100{)}^2+2(100)(1)+(1{)}^2$
5. Calculate each part:
   • ${100}^2=10000$.
   • $2(100)(1)=200$.
   • $1^2=1$.
6. Add them up: $10000+200+1=10201$.
7. So, the answer is 10201. This shows how SSC GD Algebra shortcuts work.

Example 4: Calculate ${15}^2-5^2$ using an algebraic identity.

Solution:

1. First, this looks like $a^2-b^2$. Here, $a=15$ and $b=5$.
2. Use the formula $a^2-b^2=(a+b)(a-b)$.
3. Substitute the numbers into the brackets:
   • $(15+5)\times (15-5)$
4. Solve the first bracket:
   • $15+5=20$
5. Solve the second bracket:
   • $15-5=10$
6. Multiply the results:
   • $20\times 10=200$
7. So, the answer is 200. This is a fast way to solve SSC GD Algebra problems involving squares.