2.2-Solving Equations having the Variable on both Sides

2.2-Solving Equations having the Variable on both Sides Important Formulae

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Grade 8 → Math → Linear Equations in One Variable → 2.2-Solving Equations having the Variable on both Sides

2.2 - Solving Equations having the Variable on both Sides
  • In equations with the variable on both sides, simplify both sides first.
  • Move the variable terms to one side and constant terms to the other side by adding or subtracting.
  • Example: Solve $3x + 5 = 2x + 9$
  • Step 1: Subtract $2x$ from both sides: $3x - 2x + 5 = 9$.
  • Step 2: Simplify: $x + 5 = 9$.
  • Step 3: Subtract 5 from both sides: $x = 4$.
  • Always check the solution by substituting the value of $x$ back into the original equation.

2.2 - Solving Equations having the Variable on both Sides

In this section, we will learn how to solve linear equations where the variable appears on both sides of the equation. These types of equations are quite common and involve simplifying the equation by moving all terms with the variable to one side, and constant terms to the other side. Let's go through the process step by step.

Steps to Solve Equations with Variables on Both Sides:
  1. Identify the terms with the variable: Look for terms that contain the variable (e.g., $x$, $y$, etc.) on both sides of the equation.
  2. Move all variable terms to one side: To simplify the equation, you can move all the terms with the variable to one side of the equation. This is done by adding or subtracting terms from both sides.
  3. Move constant terms to the other side: Similarly, you can move all constant terms (those without a variable) to the other side of the equation.
  4. Simplify the equation: Combine like terms and simplify both sides of the equation. The goal is to isolate the variable.
  5. Solve for the variable: Once the variable is isolated, solve for its value by performing the appropriate arithmetic operation.
Example 1:

Consider the equation:

$3x + 5 = 2x + 12$

Step 1: Move the terms with $x$ to one side and constants to the other side.

$3x - 2x = 12 - 5$

Step 2: Simplify both sides:

$x = 7$

The solution to the equation is $x = 7$.

Example 2:

Consider the equation:

$4y + 6 = 3y - 9$

Step 1: Move the variable terms to one side and the constants to the other side.

$4y - 3y = -9 - 6$

Step 2: Simplify both sides:

$y = -15$

The solution to this equation is $y = -15$.

Important Points to Remember:
  • Always perform the same operation on both sides of the equation to maintain the equality.
  • If the variable appears on both sides, move all the variable terms to one side and all constants to the other side.
  • After simplifying the equation, if you have a simple equation like $x = 7$, then that is the solution.
  • If the variable terms cancel out and you're left with a true statement (like $0 = 0$), it means the equation has infinitely many solutions.
  • If you end up with a false statement (like $5 = 0$), it means there is no solution to the equation.
Example 3 (No solution):

Consider the equation:

$2x + 3 = 2x - 4$

Step 1: Move the variable terms to one side and constants to the other side:

$2x - 2x = -4 - 3$

Step 2: Simplify both sides:

$0 = -7$

Since $0$ does not equal $-7$, this equation has no solution.

Example 4 (Infinite solutions):

Consider the equation:

$5x - 3 = 5x - 3$

Step 1: Move variable terms and constants to their respective sides:

$5x - 5x = -3 + 3$

Step 2: Simplify both sides:

$0 = 0$

Since both sides are equal, this equation has infinitely many solutions.

Solve the following equations and check your results.

1. 3x=2x+18

2. 5t–3=3t–5

3. 5x+9=5+3x

4. 4z+3=6+2z

5. 2x–1=14–x

6. 8x+4=3(x–1)+7

7. $x= \dfrac{}{} (x + 10) $

8. $\dfrac{2x}{3} + 1 = \dfrac{7x}{15} + 3$

9. $2y + \dfrac{5}{3} = \dfrac{26}{3} - y$

10. $3m = 5m - \dfrac{8}{5}$

Solution:

1. $3x = 2x + 18$

Solve: $3x - 2x = 18$
$x = 18$

2. $5t - 3 = 3t - 5$

Solve: $5t - 3t = -5 + 3$
$2t = -2$
$t = -1$

3. $5x + 9 = 5 + 3x$

Solve: $5x - 3x = 5 - 9$
$2x = -4$
$x = -2$

4. $4z + 3 = 6 + 2z$

Solve: $4z - 2z = 6 - 3$
$2z = 3$
$z = \dfrac{3}{2}$

5. $2x - 1 = 14 - x$

Solve: $2x + x = 14 + 1$
$3x = 15$
$x = 5$

6. $8x + 4 = 3(x - 1) + 7$

Solve: $8x + 4 = 3x - 3 + 7$
$8x + 4 = 3x + 4$
$8x - 3x = 4 - 4$
$5x = 0$
$x = 0$

7. $x = \dfrac{}{} (x + 10)$

Solve: $ x = x + 10$
$0 = 10$ (This is not a valid equation)

8. $\dfrac{2x}{3} + 1 = \dfrac{7x}{15} + 3$

Solve: $\dfrac{2x}{3} - \dfrac{7x}{15} = 3 - 1$
$\dfrac{10x}{15} - \dfrac{7x}{15} = 2$
$\dfrac{3x}{15} = 2$
$x = 10$

9. $2y + \dfrac{5}{3} = \dfrac{26}{3} - y$

Solve: $2y + y = \dfrac{26}{3} - \dfrac{5}{3}$
$3y = \dfrac{21}{3}$
$y = \dfrac{7}{3}$

10. $3m = 5m - \dfrac{8}{5}$

Solve: $3m - 5m = - \dfrac{8}{5}$
$-2m = - \dfrac{8}{5}$
$m = \dfrac{4}{5}$