SKYLINE UNIVERSITY COLLEGE
Bridge Program - MATHEMATICS (MBR1001) Student Name: Youse F Fareed Student ID: \( \quad 14324 \)
Solve each of the \( \mathbf{2 0} \) questions by choosing the correct from the given answers:
Solve \( 5 x+2=-8 \)
A \( x=-2 \)
B \( x=-1.2 \)
C \( x=1.2 \)
D \( x=2 \)
2
Solve \( (x+2)(2 x-1)=0 \)
A \( x=0 \)
B \( x=2 \) or \( -1 / 2 \)
\( C x=-2 \) or \( 1 / 2 \)
D \( x=-1 \) or 1
3 Solve \( \frac{3 x}{x-2}=\frac{3 x+10}{x} \)
A \( x=4 \)
\( B x=5 \)
C \( x=2 \) or \( -3 \frac{1}{3} \)
D \( x=1 \frac{1}{9} \)
4 Solve \( \frac{2}{3} x+5=-7 \)
A \( x=-18 \)
\( B x=-3 \)
\( C x=3 \)
D \( x=18 \)
5
Solve \( 3 x+2> 8 \)
\( A x> 10 / 3 \)
\( B \times< 2 \)
\( C x< -2 \)
D \( x> 2 \)

Step 1 ：First, we will solve \(5x + 2 = -8\). To solve for \(x\), we can subtract 2 from both sides and then divide the resulting equation by 5.

Step 2 ：\(5x = -10\)

Step 3 ：\(x = -2\)

Step 4 ：For the second equation \((x + 2)(2x - 1) = 0\), we can solve for \(x\) by setting each factor equal to zero.

Step 5 ：\(x + 2 = 0\) or \(2x - 1 = 0\)

Step 6 ：\(x = -2\) or \(x = \frac{1}{2}\)

Step 7 ：For the third equation \(\frac{3x}{x - 2} = \frac{3x + 10}{x}\), we can solve for \(x\) by cross multiplying and then simplifying the resulting equation.

Step 8 ：\(3x^2 = (3x + 10)(x - 2)\)

Step 9 ：\(3x^2 = 3x^2 - 6x + 10x - 20\)

Step 10 ：\(-16x = -20\)

Step 11 ：\(x = \frac{5}{4}\)

Step 12 ：For the fourth equation \(\frac{2}{3}x + 5 = -7\), we can solve for \(x\) by subtracting 5 from both sides and then multiplying the resulting equation by \(\frac{3}{2}\).

Step 13 ：\(\frac{2}{3}x = -12\)

Step 14 ：\(x = -18\)

Step 15 ：For the fifth inequality \(3x + 2 > 8\), we can find the values of \(x\) that satisfy the inequality by subtracting 2 from both sides and then dividing the resulting inequality by 3.

Step 16 ：\(3x > 6\)

Step 17 ：\(x > 2\)