High School

7. Find x in the following:

(i) 3 x [(-6) + x] = 3 x (6) + 3 x 10
(iii) x x [6 + (-8)] = 15 x 6 + (-8)×15

Answer :

To solve for [tex]x[/tex] in the given equations, we will handle each equation separately.

Equation (i):

[tex]3 \times [(-6) + x] = 3 \times 6 + 3 \times 10[/tex]

First, simplify both sides of the equation:

  • Left side:
    [tex]3 \times [(-6) + x] = 3 \times -6 + 3 \times x = -18 + 3x[/tex]

  • Right side:
    [tex]3 \times 6 + 3 \times 10 = 18 + 30 = 48[/tex]

Now equate the simplified sides:

[tex]-18 + 3x = 48[/tex]

Add 18 to both sides to isolate the term with [tex]x[/tex]:

[tex]3x = 48 + 18[/tex]

[tex]3x = 66[/tex]

Divide both sides by 3 to solve for [tex]x[/tex]:

[tex]x = \frac{66}{3} = 22[/tex]

So, [tex]x = 22[/tex] for equation (i).

Equation (iii):

[tex]x \times [6 + (-8)] = 15 \times 6 + (-8) \times 15[/tex]

First, simplify both sides of the equation:

  • Inside the brackets on the left side:
    [tex]6 + (-8) = -2[/tex]

  • Therefore, the left side becomes:
    [tex]x \times (-2) = -2x[/tex]

  • Simplify the right side:
    [tex]15 \times 6 = 90[/tex]
    [tex](-8) \times 15 = -120[/tex]

  • Combine the results:
    [tex]90 + (-120) = -30[/tex]

Equating both sides:

[tex]-2x = -30[/tex]

Divide both sides by -2 to solve for [tex]x[/tex]:

[tex]x = \frac{-30}{-2} = 15[/tex]

So, [tex]x = 15[/tex] for equation (iii).

Therefore, the solutions are [tex]x = 22[/tex] for equation (i) and [tex]x = 15[/tex] for equation (iii).

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