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Standard VII

Mathematics

Multiplication of Integers

Now, can't yo...

Question

Now, can’t you compute these differences?

(i) 128 − 176

(ii) 248 − 312

(iii)

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Solution

We know that for any two positive numbers x and y, with x < y, we have x − y = −(y − x)

(i) 128 − 176 = −(176 − 128) = −48

(ii) 248 − 312 = −(312 − 248) = −64

(iii) = =

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Similar questions

Now can’t you compute these?

(i) −4 − (−5)

(ii) −6 − (−9)

(iii) −12 − (−10)

(iv) −8 − (−7)

(v)

Now can’t you do these?

(i) −4 + (−1)

(ii) −6 + (−4)

(iii) −12 + (−8)

(iv) −8 + (−10)

(v)

Now can’t you do these?

(i) −3 −1

(ii) −5 −6

(iii) −8 −8

(iv) −12 −18

(v)

Now compute these.

(i) $- 4 + 1$

(ii) $- 5 + 3$

(iii) $- 9 + 2$

(iv) $- 12 + 8$

(v) $- 14 + 1$

(vi) $- \frac{3}{4} + \frac{1}{4}$

Now can’t you simplify these?

(i) 2x + (3x + 1)

(ii) x + (2x + 3)

(iii) (x + 2) + (2x + 1)

(iv) (x + 1) + (x + y)

(v) (3x + 2y) + (x + y)

(vi) (3x + 2y) + (2x + 3y)

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Now, can’t you compute these differences? (i) 128 − 176 (ii) 248 − 312 (iii)

We know that for any two positive numbers x and y, with x < y, we have x − y = −(y − x) (i) 128 − 176 = −(176 − 128) = −48 (ii) 248 − 312 = −(312 − 248) = −64 (iii) = =

Now, can’t you compute these differences?

(i) 128 − 176

(ii) 248 − 312

(iii)

We know that for any two positive numbers x and y, with x < y, we have x − y = −(y − x)

(i) 128 − 176 = −(176 − 128) = −48

(ii) 248 − 312 = −(312 − 248) = −64

(iii) = =