Place the last two terms of the following expressions in parentheses preceded by a minus sign:
(i) x + y − 3z + y
(ii) 3x − 2y − 5z − 4
(iii) 3a − 2b + 4c − 5
(iv) 7a + 3b + 2c + 4
(v) 2a² − b² − 3ab + 6
(vi) a² + b²− c² + ab − 3ac

Open in App

Solution

We have
(i) x + y − 3z + y = x + y − (3z - y )
(ii) 3x − 2y − 5z − 4 = 3x - 2y - (5z + 4)
(iii) 3a − 2b + 4c − 5 = 3a - 2b - (- 4c + 5)
(iv) 7a + 3b + 2c + 4 = 7a + 3b - (- 2c - 4)
(v) 2a² − b² − 3ab + 6 = 2a² − b² − (3ab - 6)
(vi) a²+ b² − c² + ab − 3ac = a²+ b² − c² - (- ab + 3ac)

Suggest Corrections

Similar questions

3 a - 2 b + 4 c - 5

\frac{(b^{2} + c^{2})}{3 b c} + \frac{(c + a)^{2}}{3 a c} + \frac{(a + b)^{2}}{3 a b} = \frac{1}{3}

27 a^{3} + b^{3} + c^{3} - 9 a b c = (3 a + b + c) [9 a^{2} + b^{2} + c^{2} - 3 a b - b c - 3 a c]

S

⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦

b_{1}, b_{2}, b_{3} \in R

\begin{matrix} - x + 2 y + 5 z = b_{1} 2 x - 4 y + 3 z = b_{2} x - 2 y + 2 z = b_{3} \end{matrix}

⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦ \in S

a + b + c = 0

\frac{(b + c)^{2}}{3 b c} + \frac{(c + a)^{2}}{3 a c} + \frac{(a + b)^{2}}{3 a b}

Join BYJU'S Learning Program