<h3>1. <u>Answer;</u></h3>
<em>p =-3/2p</em>
<h3><u>Solution;</u></h3>
To solve for P in the equation;
= 3(p + q ) = p
We open the brackets;
= 3p + 3q = p
Then, we combine the like terms together, we get;
<em> 3p - p = -3q</em>
<em> Therefore; 2p = -3q</em>
<em> hence; p = -3/2q. </em>
<h3>2. <u>
Answers</u>;</h3>
<em>b = 19/2</em>
<h3><u>Solution;</u></h3>
To solve for b when the value of a = 3
4a = 2b - 7
we substitute 3 with a in the equation
<em>( 4 × 3 ) = 2b -7</em>
<em>12 = 2b -7</em>
<em>Then, make b the subject</em>
<em>19 = 2b</em>
<em>b = 19/2</em>
<h3>3. <u>Answer;</u></h3>
<em>r =d/t</em>
<h3><u>
<em>Solution;</em></u></h3>
To solve for r
d= rt.
In order to get r, we divide both sides by t
<em>d/t = rt/t</em>
<em>r = d/t</em>
<h3>4. <u>Answer;</u></h3>
<em>Width = 30 units</em>
<h3><u>Solution;</u></h3>
The width of a rectangle with a given perimeter of 90 and length is 15.
The perimeter of a rectangle is given by;
Perimeter= 2(Length + Width)
Thus;<em> Perimeter = 90 , length=15 </em>
<em> 90 = 2 (W + 15)</em>
<em>Dividing both sides by 2,</em>
<em>45 = w + 15 </em>
<em>Subtracting from both sides;</em>
<em>W = 45 -15 </em>
<em>W = 30 </em>
Therefore,<em><u> the width of the rectangle is 30 units</u></em>
