If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y)
3
2
1
5
Correct answer is A
\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)
\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2)
x + 4y = 6......(iii)
3x - 4y = 2.....(iv) add (iii) and (iv)
4x = 8, x = \(\frac{8}{4}\) = 2
substitute x = 2 into equation (iii)
x + 4y = 6
2 + 4y = 6
4y = 6 - 2
4y = 4
y = \(\frac{4}{4}\)
= 1(x + y)
2 + 1 = 3
If x = 64 and y = 27, evaluate: \(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)
2\(\frac{1}{5}\)
1
\(\frac{5}{11}\)
\(\frac{11}{43}\)
Correct answer is C
\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)
substitute x = 64 and y = 27
\(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\)
= \(\frac{8 - 3}{27 - 16}\)
= \(\frac{5}{11}\)
Solve the inequality: \(\frac{2x - 5}{2} < (2 - x)\)
x > 0
x < \(\frac{1}{4}\)
x > 2\(\frac{1}{2}\)
x < 2\(\frac{1}{4}\)
Correct answer is D
\(\frac{2x - 5}{2} < \frac{(2 - x)}{1}\)
2x - 5 < 4 - 2x
2x + 2x < 4 + 5
4x < 9
x < \(\frac{9}{4}\)
x < 2\(\frac{1}{4}\)
Simplify: \(\frac{x^2 - y^2}{(x + y)^2} \div \frac{(x - y)^2}{(3x + 3y)}\)
\(\frac{x - y}{3}\)
x + y
\(\frac{3}{x - y}\)
x - y
Correct answer is C
\(\frac{x^2 - y^2}{(x + y)^2} \div \frac{(x - y)^2}{(3x + 3y)}\)
\(\frac{(x + y)(x - y)}{(x + y)(x + y)}\div \frac{(x - y)(x - y)}{3(x + y)}\)
= \(\frac{3}{x - y}\)
N94x + 6y)
N(6x + 4y)
N(24x + 12y)
N(12x + 24y)
Correct answer is B
4 oranges sell for Nx, 1 orange will sell for \(\frac{Nx}{4}\)
24 oranges will sell for: \(\frac{Nx}{4} \times 24\) = n6x
3 mangoes sell for Ny, 1 mango will sell for \(\frac{Ny}{3}\)
12 mangoes will sell for \(\frac{Ny}{3} \times 12\) = 4Ny
total money pay N6x + N4y = N(6x + 4y)