Math 103 – Intermediate Algebra
Test 1 Review Solutions

1. 	a. 	m = -2                                                    	b.	m = 1/3 

c. 	y = 2x   m = 2	d. 	y =    m = 1/2

2.	a.	m for both is 1/7; so the lines 	b. 	m = 2, m = -1/2; the slopes are			are parallel.		negative reciprocals so the lines are
				perpendicular.	      

	c.	, 	d.	, , m is 4 for	
	m = 3/2, m = -2/3; the slopes are 		both, so the lines are parallel.
	negative reciprocals so the lines are 
	perpendicular. 
					
3	a.	i) 	m parallel is  
		    Conclusion: 

		 ii) 	m perpendicular is 3 
		        Conclusion: 

	b.	 i) 	m parallel is 2
 		  Conclusion: 
   		ii) 	m perpendicular is 
 		  Conclusion: 

4. 	a.		b.	 

Check: 	Check:  
                                    
	c.		d.	

Check:  = 4(3)		       
                          	Check:   ?    
             			    ?
                 			           yes

5.	a.	S = C + rC	b.	h = s0 +  v0t  at2
     		S = C(1 + r)		2h = 2s0 +  2v0t  at2
				2h  2s0 +  at2 = 2v0t 
				
	c.	ax + by = c
		by =  ax + c

6.	a.	an identity	b.	a conditional equation  

	c.		d.    
		an equation with no solution		an identity			

7.	a.	 	Check: 
 		4x + 2 = 10 or 4x + 2 = 10               x = 2:?    x = 3:?    
		4x  = 8             4x =  12	   ?	          ?
		   x = 2               x  =  3                              | 10 | = 10, yes               |10 | = 10 yes              

	b.	
		
		8y – 32 = 2y + 6 or 8y – 32 = 2y  6
       	 6y = 38                 10y = 26
       	y =        y = 

Check: y = :        	y = :  ?                       
                         ?	            ?
                                  ?	              	   ?
                                  , yes                                          , yes      
	c.	 - no solution, an absolute value is never negative.

	d.   	Check: : ?
		 or 	            	 ?  , yes.
		  7a + 3 = 8         7a + 3 =  8				
                   7a = 5                 7a =  11		   : ?
                                                                      ? yes.
					            
	e.                         	Check: x = 21: ?      
                                                                 3 | 7 | =  21?    3(7)  =  21, yes          

	          x – 14 = 7 or   x – 14 = – 7                                                
                      x = 21               x = 7                                            x = 7: ?
                                                                                  3 |  7 | =  21?  3(7)  =  21, yes  
8.	a. 3 – 5 < 2 – 3(3)?	b.	| 3 – 7 | > 2
	    – 2 < 2 – 9?		| – 4 | > 2
	    – 2 < – 9 no, 		    4  > 2, yes
x = 3 is not a solution to the inequality.	x = 3 is a solution to the inequality.

9.	a.	     	b.   8x + 8 < 7x + 35
         3(x – 4) > 2(x – 2) + 5	     8x + 8  <   7x + 35 
          3x – 12 > 2x – 4 + 5	  – 7x          – 7x____  
          3x – 12  >  2x + 1                                    		 x  + 8 <            35
	 – 2x           – 2x                                                          – 8              – 8
           x – 12  >     1                                                      x    <                27             
           + 12      + 12                                              
		    x  >   13                                               

	c. 		d.	4x – 3 > 5
		6  <  8 – x  <  26		4x > 8
         – 8   – 8          – 8		 x > 2
         – 2  <     – x  < 18 		and
	(–1)( –2) > (–1)( –x) > (–1)(18)		4 – x > 3
       	 2 > x > –18   		4 > 3 + x
				1 > x
	x < 1 and x > 2, impossible, no solution

10. a.		b.	
 | 2x – 2 + 4| < 8  | 2x + 2 | < 8		 x – 4 < –2 or x – 4 > 2
                          8 < 2x + 2 < 8		x < 2  or x > 6                                         
                           10 < 2x < 6	
                              5 < x < 3             	

	c.	Every real number has an absolute value greater than  2.

     
11. Let C be the price of the computer: C + .06C = 1187.2
		                                                       1.06C = 1187.2

12.	Alida’s rate = waking rate + moving sidewalk rate = 4 + 5 = 9ft/sec
	Distance = ratetime  300 = 9t  t =  33.33seconds
13.	Area = baseheight  72 = b6  b = 72/6 = 12cm

14.	Let t be the length of time it takes them to meet.
	Distance = ratetime, so the distance the first car has driven after t hours is 55t and the distance for the second car is 50t. The total distance is 157.5 miles so:
	55t + 50t = 157.5105t = 157.5
	                             t = 157.5/105 = 1.5 hours

15.	Let x be the amount of 1% milk that is added. 
	Then the total amount of the new mixture will be 12 + x
	The amount of butter fat in any part of the mixture is given by the percent times 	the amount. We want the total amount to be 2% times total amount: .02(12 + x)
	The amount of butter fat already there is .04(12), The amount we are adding is .01x
	equation: .04(12) + .01x = .02(12 + x) .48 + .01x = .24 + .02x
multiply by 100 to clear decimals:  48 + x =  24 + 2x
subtract 24 and x from both sides:    24 = x    answer: add 24 liters of 1% milk.

16. If x is the length, the perimeter is 2x + 2w = 2x + 2(8) = 2x + 16
	So 32 < 2x + 16 < 48.
		16 < 2x < 32
		8 < x < 16

17. a.	Find the equation of the line through points (45, 10) and (20, 60).
	
	p – 10 = –2(x – 45)  p – 10 = –2x + 90  2x = – p + 100   x = 	

	b.	When p = 5, x = = 47.5, so the demand would be about 47 		or 48 phones.