Factorize; (2x + 3y)2 - (x - 4y)2
(3x - y)(x + 7y)
(3x + y)(2x - 7y)
(3x + y)(x - 7y)
(3x - y)(2x + 7y)
Correct answer is A
(2x + 3y)2 - (x - 4y)2
= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)
= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)
= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2
= 3x2 + 20xy - 7y2
= 3x2 + 21xy - xy - 7y2
= 3x(x + 7y) - y(x + 7y)
= (3x - y)(x + 7y)
Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)
s = \(\frac{mrp}{nr + m^2}\)
s = \(\frac{nr + m^2}{mrp}\)
s = \(\frac{nrp}{mr + m^2}\)
s = \(\frac{nrp}{nr + m^2}\)
Correct answer is D
P = S + \(\frac{sm^2}{nr}\)
P = S(1 + \(\frac{m^2}{nr}\))
P = S(1 + \(\frac{nr + m^2}{nr}\))
nrp = S(nr + m2)
S = \(\frac{nrp}{nr + m^2}\)
Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\)
\(\frac{x + 1}{x(1 - x)}\)
\(\frac{3x - 1}{ x(1 - x)}\)
\(\frac{3x + 1}{ x(1 - x)}\)
\(\frac{x + 1}{ x(1 - x)}\)
Correct answer is D
\(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\)
= \(\frac{2x - 1(1 + x)}{x(1 - x)}\)
= \(\frac{3x - 1}{x(1 - x)}\)
Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?
1
3
7
17
Correct answer is C
2x + y = 7...(1)
3x - 2y = 3...(2)
From (1), y = 7 - 2x for y in (2)
3x - 2(7 - 2x) = 3
3x - 14 + 4x = 3
7x + 3 + 14 = 17
x = \(\frac{17}{7}\)
Hence, 7 x \(\frac{17}{7}\)
= 17 - 10
= 7
Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined
6, -7
3, -6
3, -7
-3, -7
Correct answer is C
\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\)
Factorize the denominator;
Y2 + 7y - 3y - 21
= y(y + 7) -3 (y + 7)
= (y - 3)(y + 7)
Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined
when y2 + 4y - 21 = 0
ie. y = 3 or -7