Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

By eliminating variable 'z' as there are three unknowns & only 2 equations.
By putting z = 0
3x + 5y = 49 . . . . . . (i)
9x + 8y = 126 . . . . . . (ii)
Multiplying by 3 in equation (i) and subtracting equation (ii)
9x + 15y = 147
9x + 8y = 126
$$\overline {\,\,\,\,\,\,\,\,\,\,\,\,7{\text{y}} = 21\,\,} $$
y = 3

Alternate solution:
3x + 5y + 7z = 49
9x + 8y + 21z = 126
Assume value x, y, z
x = 2, y = 3, z = 4
3(2) + 5(3) + 7(4) = 49
6 + 15 + 28 = 49
49 = 49 value satisfied
9(2) + 8(3) + 21(4) = 126
18 + 24 + 84 = 126
126 = 126 value satisfied
y = 3

x² + 8y² + 12y - 4xy + 9 = 0
x² + 4y² - 4xy + 4y² + 12y + 9 = 0
(x - 2y)² + (2y + 3)² = 0
x = 2y
2y = -3
y = $$\frac{{ - 3}}{2}$$
x = -3
(7x + 8y)
= 7 × (-3) + 8 × $$\left( {\frac{{ - 3}}{2}} \right)$$
= -21 - 12
= -33

(x + y)³ + 8(x - y)³ = (3x + Ay)(3x² + Bxy + Cy²)
a³ + b³ = (a + b)(a² - ab + b²)
a = (x + y)
b = 2(x - y)
(x + y)³ + 8(x - y)³ = (3x - y)(x² + y² + 2xy - 2x² + 2y² + 4x² + 4y² - 8xy)
(x + y)³ + 8(x - y)³ = (3x - y)(3x² - 6xy + 7y²)
(3x + Ay)(3x² + Bxy + Cy²) = (3x - y)(3x² - 6xy + 7y²)
Compare
A = -1
B = -6
C = 7
A + B + C = -1 - 6 + 7 = 0

$$\eqalign{ & {\text{Let }}\sqrt {\frac{a}{b}} = x \cr & \therefore \,x - 13 = \frac{{ - 1}}{x} - 11 \cr & x + \frac{1}{x} = 2 \cr & \therefore \,x = 1 \cr & \sqrt {\frac{a}{b}} = 1{\text{ and }}a + b = 10 \cr & \therefore \,a = b = 5 \cr & 3ab + 4{a^2} + 5{b^2} \cr & = 3{a^2} + 4{a^2} + 5{a^2} \cr & = 12{a^2} \cr & = 12 \times 25 \cr & = 300 \cr} $$

16x² + 9y² + 4z² = 24(x - y + z) - 61
⇒ 16x² + 9y² + 4z² = 2 × 4x × 3 - 2 × 3x × 4 + 2 × 2z × 6
⇒ (4x - 3)² + (3y + 4)² + (2z - 6)² = 0
x = $$\frac{3}{4}$$
y = $$ - \frac{4}{3}$$
z = 3
⇒ xy + 2z
$$\eqalign{ & = \frac{3}{4} \times \left( { - \frac{4}{3}} \right) + 2 \times 3 \cr & = - 1 + 6 \cr & = 5 \cr} $$

Let a = b = c = 1
∴ x = 2, y = 2, z = 2
xyz - x² - y² - z²
= 2 × 2 × 2 - 4 - 4 - 4
= 8 - 12
= -4

$$\eqalign{ & \because \,{x^2} + \frac{1}{{25{x^2}}} = \frac{8}{5} \cr & \Rightarrow {x^2} + \frac{1}{{25{x^2}}} + 2x \times \frac{1}{{5x}} = \frac{8}{5} + \frac{2}{5} \cr & \Rightarrow {\left( {x + \frac{1}{{5x}}} \right)^2} = 2 \cr & \Rightarrow x + \frac{1}{{5x}} = \sqrt 2 \cr & {\text{On cubing both sides}} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} + 3 \times \frac{1}{5}\left( {\sqrt 2 } \right) = {\left( {\sqrt 2 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = 2\sqrt 2 - \frac{{3\sqrt 2 }}{5} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = \frac{{7\sqrt 2 }}{5} \cr} $$

It intersect at point P(a, b) = (x, y)
3x - 20y = 2
11x - 5y = -61
44x - 20y = -244
3x - 20y = 2
$$\overline {41{\text{x}}\,\,\,\,\,\,\,\,\, = - 246\,\,} $$
x = -6
y = -1
P(a, b) = (-6, -1)
$$\eqalign{ & \frac{{{a^2} + {b^2} - ab}}{{{a^2} - {b^2} + ab}} \cr & = \frac{{36 + 1 - 6}}{{36 - 1 + 6}} \cr & = \frac{{31}}{{41}} \cr} $$

$$\eqalign{ & {\text{We have,}} \cr & \Rightarrow a + b + c = \frac{7}{{12}}\,......\,\left( 1 \right) \cr & \Rightarrow 3a - 4b + 5c = \frac{3}{4}\,......\,\left( 2 \right) \cr & \Rightarrow 7a - 11b - 13c = - \frac{7}{{12}}\,......\,\left( 3 \right) \cr} $$
Subtracting 3 times of equation (1) from equation (2), we get
$$\eqalign{ & \Rightarrow - 7b + 2c = \frac{3}{4} - 3 \times \frac{7}{{12}} \cr & \Rightarrow - 7b + 2c = \frac{3}{4} - \frac{7}{4} \cr & \Rightarrow - 7b + 2c = - 1\,\,......\,\left( 4 \right) \cr} $$
Subtracting equation (3) from 7 times equation (1), we get
$$\eqalign{ & \Rightarrow 18b + 20c = \frac{{49}}{{12}} + \frac{7}{{12}} \cr & \Rightarrow 18b + 20c = \frac{{56}}{{12}} \cr & \Rightarrow 18b + 20c = \frac{{14}}{3}\,......\,\left( {\text{5}} \right) \cr} $$
Subtracting 10 times of equation (4) from equation (5), we get
$$\eqalign{ & \Rightarrow 88b = \frac{{14}}{3} + 10 \cr & \Rightarrow 88b = \frac{{44}}{3} \cr & \Rightarrow b = \frac{1}{6} \cr} $$
Substituting in equation (1), we get
$$\eqalign{ & \Rightarrow a + \frac{1}{6} + c = \frac{7}{{12}} \cr & \Rightarrow a + c = \frac{7}{{12}} - \frac{1}{6} = \frac{5}{{12}} \cr & \therefore {\text{The value of }}a + c{\text{ is }}\frac{5}{{12}} \cr} $$

(2x + 3)³ + (x - 8)³ + (x + 13)³ = 3(2x + 3)(x - 8)(x + 13)
a³ + b³ + c³ - 3abc = 0   {If a + b + c = 0}
2x + 3 + x - 8 + x + 13 = 0
4x + 8 = 0
x = -2