Tag: use of brackets

Questions Related to use of brackets

$ x + y -(z - x -[y + z - (x + y - \left { z + x - (y + z + x) \right})])$ is equal to

maths use of brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. 3x

  2. 2y

  3. x

  4. 0

Show answer
Correct Option: C
Explanation:

On simplifying, we get
$x + y -(z - x -[y + z - (x + y - \left { z + x - (y + z + x) \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y - \left { z + x - y - z - x \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y - \left { - y \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y + y)])\$ 
$= x + y -(z - x -[y + z - x - y - y)])\$ 
$= x + y -(z - x -[ z - x - y])\$ 
$= x + y -(z - x - z + x + y)\$ 
$= x + y -(y)\$ 
$= x.$

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = $

maths use of brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. $-\frac{4}{3}p$

  2. 5q

  3. -7r

  4. 5q-7r

Show answer
Correct Option: D
Explanation:

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = \frac{1}{3}(-2p+6q-9r)-\frac{1}{3}(-2p -9q +12r) $
$=\frac{1}{3} [\left (-2p+6q-9r)-(-2p -9q +12r) \right]$
$=\frac{1}{3} (-2p + 6q - 9r + 2p + 9q -12r)$
$=\frac{1}{3} (15q - 21r)$
$=(5q - 7r)$

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$

maths use of brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. $(a+y)^{2}$

  2. $\frac{3a}{16}$

  3. $\frac{9}{16}(a+y)^{2}$

  4. 1

Show answer
Correct Option: C
Explanation:

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left (\frac{3}{4}(a+y) \right) \right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{4}(a+y)\right ]$
= $ \frac{3}{4}(a+y) \left [ \frac{3}{4}(a+y)\right ]$
= $ \frac{9}{16}(a+y)^2$