How do you subtract $3 x^{2} - 4$ from the sum of $x^{2} - 6 x + 1$ and $2 x^{2} - 5 x + 5$ ?

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Solution

The sum of $x^{2} - 16 x + 1$ and $2 x^{2} - 5 x + 5$ is solved as:
$x^{2} - 6 x + 1 + 2 x^{2} - 5 x + 5 \Rightarrow$
$x^{2} + 2 x^{2} - 6 x - 5 x + 1 + 5 \Rightarrow$
$3 x^{2} - 11 x + 6 \Rightarrow$
Then to subtract $3 x^{2} - 4$ from this is solved as:
$3 x^{2} - 11 x + 6 - (3 x^{2} - 4) \Rightarrow$
$3 x^{2} - 11 x + 6 - 3 x^{2} + 4) \Rightarrow$
$3 x^{2} - 3 x^{2} - 11 x + 6 + 4$
$- 11 x + 10$ .

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Question 6 (b)

From the sum of $4 + 3 x a n d 5 - 4 x + 2 x^{2}$ , subtract the sum of $3 x^{2} - 5 x a n d - x^{2} + 2 x + 5$ .

From the sum of $4 + 3 x$ and $5 - 4 x + 2 x^{2}$ , subtract the sum of $3 x^{2} - 5 x$ and $- x^{2} + 2 x + 5$ .

4 + 3 x

5 - 4 x + 2 x^{2}

3 x^{2} - 5 x

- x^{2} + 2 x + 5

(a) From the sum of 3x − y + 11 and − y− 11, subtract 3x − y − 11.

(b) From the sum of 4 + 3x and 5 − 4x + 2x²,subtract the sum of 3x² − 5x and
−x² + 2x+ 5.

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