Equations in One Variable - Grade 6

An equation in mathematics is a statement that two mathematical expressions are equal. Therefore, an equation must contain an equal sign (=).

From the given list, only the following contain an equal sign and are therefore equations:

• b) \(x + 2 = 4\)
• c) \(8 = x\)
• e) \(\dfrac{x - 8}{3} = 9\)

The others (a, d, f) are just mathematical expressions.

According to the definition, an equation must have an equal sign. The following are not equations (they are expressions):

• a) \(3x - 9\)
• d) \(8 \times x\)

Note: Option c (\(12 - 7 = 5\)) is a numerical equation, and options b and e are algebraic equations.

We test the values of \(x\) by substituting them into the equation \(\;2x - 4 = 4\; \) and comparing both sides.

• Testing \(x = 0\):
  Left side: \(2(0) - 4 = -4\)
  Right side: \(4\)
  \(-4 \neq 4\) → Not a solution.
• Testing \(x = 4\):
  Left side: \(2(4) - 4 = 8 - 4 = 4\)
  Right side: \(4\)
  \(4 = 4\) → \(x = 4\) is the correct solution.

Check which values satisfy \(\dfrac{x}{3} - 1 = 2\):

• a) \(x = -3\): Left side = \(\dfrac{-3}{3} - 1 = -1 - 1 = -2\). Right side = \(2\). Not a solution.
• b) \(x = 6\): Left side = \(\dfrac{6}{3} - 1 = 2 - 1 = 1\). Right side = \(2\). Not a solution.
• c) \(x = -9\): Left side = \(\dfrac{-9}{3} - 1 = -3 - 1 = -4\). Right side = \(2\). Not a solution.
• d) \(x = 9\): Left side = \(\dfrac{9}{3} - 1 = 3 - 1 = 2\). Right side = \(2\). This is the solution.

To solve an equation, we isolate \(x\) by applying the inverse operation to both sides of the equal sign:

1. \(x - 6 = 12\)
   Add 6 to both sides: \(x = 12 + 6 \Rightarrow \mathbf{x = 18}\)
2. \(3 = x + 3\)
   Subtract 3 from both sides: \(3 - 3 = x \Rightarrow \mathbf{x = 0}\)
3. \(2 + x = 8\)
   Subtract 2 from both sides: \(x = 8 - 2 \Rightarrow \mathbf{x = 6}\)
4. \(2x = 16\)
   Divide both sides by 2: \(x = \dfrac{16}{2} \Rightarrow \mathbf{x = 8}\)
5. \(\dfrac{x}{3} = 5\)
   Multiply both sides by 3: \(x = 5 \times 3 \Rightarrow \mathbf{x = 15}\)

We solve both equations in each pair and compare their solutions:

1. \(x = 2\) and \(2x = 4\) (which simplifies to \(x=2\))
   Same solutions.
2. \(x + 3 = 6 \Rightarrow x = 3\)
   \(x + 4 = 8 \Rightarrow x = 4\)
   Different solutions.
3. \(\dfrac{x}{2} = 2 \Rightarrow x = 4\)
   \(x = -4\)
   Different solutions.
4. \(3x = 9 \Rightarrow x = 3\)
   \(x + 1 = 4 \Rightarrow x = 3\)
   Same solutions.

Pairs (a) and (d) have the same solutions.

Set up the equation: \(2x + 6 = 12\)

Subtract 6 from both sides: \(2x = 6\)

Divide by 2: \(\mathbf{x = 3}\)

Check: \(2(3) + 6 = 6 + 6 = 12\) ✔

Set up the equation: \(4x + 6 = 2 + 12\)

Simplify the right side: \(4x + 6 = 14\)

Subtract 6 from both sides: \(4x = 8\)

Divide by 4: \(\mathbf{x = 2}\)

Check: \(4(2) + 6 = 8 + 6 = 14\) and \(2 + 12 = 14\) ✔

Translate "sum" to addition: \(d + 23 = 56\)

Subtract 23 from both sides: \(d = 56 - 23\)

\(\mathbf{d = 33}\)

Translate "subtracted from" carefully (the 7 comes after the x): \(x - 7 = 41\)

Add 7 to both sides: \(x = 41 + 7\)

\(\mathbf{x = 48}\)

Translate "product" to multiplication: \(6y = 36\)

Divide both sides by 6: \(y = \dfrac{36}{6}\)

\(\mathbf{y = 6}\)

Translate to an equation: \(\dfrac{b}{5} = 4\)

Multiply both sides by 5: \(b = 4 \times 5\)

\(\mathbf{b = 20}\)

Set up the equation representing their total cards: \(x + 23 = 121\)

Subtract 23 from both sides: \(x = 121 - 23\)

\(\mathbf{x = 98}\) (Jacky has 98 cards)

Let \(c\) be Toby's contribution. Set up the equation: \(34 + 45 + c = 123\)

Simplify the constants on the left side: \(79 + c = 123\)

Subtract 79 from both sides: \(c = 123 - 79\)

\(\mathbf{c = 44}\) (Toby contributed $44)