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(a) Simplify $(6x-3)(2x-1)$ - Junior Cycle Mathematics - Question Question 1

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(a) Simplify $(6x-3)(2x-1)$. (b) Simplify $(3x^2 - 2x^3 - 3x + 2)(x - 1)$. (c) (i) Solve the simultaneous equations: $2x - 3y = 18$ $5x + 9y = -10$. (ii) V... show full transcript

Worked Solution & Example Answer:(a) Simplify $(6x-3)(2x-1)$ - Junior Cycle Mathematics - Question Question 1

Step 1

Simplify $(6x-3)(2x-1)$

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Answer

To simplify the expression, we apply the distributive property:

(6x3)(2x1)=6x(2x)6x(1)3(2x)+3(1).(6x-3)(2x-1) = 6x(2x) - 6x(1) - 3(2x) + 3(1).

Calculating each term gives:

  • 12x212x^2
  • 6x-6x
  • 6x-6x
  • +3+3

Combining like terms results in:

12x212x+3.12x^2 - 12x + 3.

Thus, the simplified form is:

Answer: 12x212x+312x^2 - 12x + 3

Step 2

Simplify $(3x^2 - 2x^3 - 3x + 2)(x - 1)$

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Answer

First, distribute each term in (3x22x33x+2)(3x^2 - 2x^3 - 3x + 2) by (x1)(x - 1):

  • For 3x23x^2, we get:
    3x2imesx3x2imes1=3x33x2,3x^2 imes x - 3x^2 imes 1 = 3x^3 - 3x^2,
  • For 2x3-2x^3, we have: 2x3imesx(2x3)imes1=2x4+2x3,-2x^3 imes x - (-2x^3) imes 1 = -2x^4 + 2x^3,
  • For 3x-3x, we find: 3ximesx+3ximes1=3x2+3x,-3x imes x + 3x imes 1 = -3x^2 + 3x,
  • For 22, we can calculate: 2imesx2imes1=2x2.2 imes x - 2 imes 1 = 2x - 2.

Combining all these results:

2x4+(3x3+2x3)(3x2+3x2)+(2x+3x)2.-2x^4 + (3x^3 + 2x^3) - (3x^2+3x^2) + (2x + 3x) - 2.

Simplifying gives: 2x4+5x36x2+5x2. -2x^4 + 5x^3 - 6x^2 + 5x - 2.

Thus, the answer is:

Answer: 2x4+5x36x2+5x2-2x^4 + 5x^3 - 6x^2 + 5x - 2

Step 3

Solve the simultaneous equations: $2x - 3y = 18$ and $5x + 9y = -10$

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Answer

To solve these equations, we can use substitution or elimination. Here, we will use substitution:

Starting with the first equation: 2x3y=182x - 3y = 18 Rearranging gives:

\Rightarrow x = \frac{3y + 18}{2}.$$ Now, substituting this into the second equation: $$5\left( \frac{3y + 18}{2} \right) + 9y = -10.$$ Multiply everything by 2 to eliminate the fraction: $$5(3y + 18) + 18y = -20 \ 15y + 90 + 18y = -20 \ 33y + 90 = -20 \ 33y = -20 - 90 \ 33y = -110 \ y = -\frac{110}{33} \ y = -\frac{10}{3}.$$ Substituting $y$ back to find $x$: $$x = \frac{3(-\frac{10}{3}) + 18}{2}\ \Rightarrow x = \frac{-10 + 18}{2}\ \Rightarrow x = \frac{8}{2} \Rightarrow x = 4.$$ Thus, the solution to the simultaneous equations is: **Answer: $x = 4$, $y = -\frac{10}{3}$**

Step 4

Verify your answer to (c)(i).

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Answer

To verify the solution x=4x = 4 and y=103y = -\frac{10}{3}, using the first equation:

8 + 10 = 18.$$ This holds true. Now, using the second equation: $$5(4) + 9(-\frac{10}{3}) = -10. \ 20 - 30 = -10.$$ This also holds true. Thus, we have confirmed the answers are correct. **Verification passed**

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