Identify graphs of linear inequalities on a number line.


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1 COMPETENCY 1.0 KNOWLEDGE OF ALGEBRA SKILL 1.1 Identify grphs of liner inequlities on number line.  When grphing firstdegree eqution, solve for the vrible. The grph of this solution will be single point on the number line. There will be no rrows.  When grphing liner inequlity, the dot will be hollow if the inequlity sign is < or >. If the inequlity signs is either or, the dot on the grph will be solid. The rrow goes to the right for or >. The rrow goes to the left for or<. Solve: 5( x + ) + x = 3( x) 5x x = 3x6 7x+ 10 = 3x6 4x = 16 x = Solve: (3x 7) > 10x 6x 14 > 10x 4x > 1 x < 3 Note the chnge in inequlity when dividing by negtive numbers
2 Solve the following equtions nd inequlities. Grph the solution set. 1. 5x 1> 14. 7(x 3) + 5x = 19 x 3. 3x + 4 1x ( x + 3) = 9 SKILL 1. Identify grphs of liner equtions nd inequlities in the coordinte plne. A first degree eqution hs n eqution of the form x + by = c. To grph this eqution, find either one point nd the slope of the line or find two points. To find point nd slope, solve the eqution for y. This gets the eqution in slope intercept form, y = mx + b. The point (0,b) is the y intercept nd m is the line's slope. To find ny points, substitute ny numbers for x nd solve for y. To find the intercepts, substitute 0 for x nd then 0 for y. Remember tht grphs will go up s they go to the right when the slope is positive. Negtive slopes mke the lines go down s they go to the right. If the eqution solves to x = ny number, then the grph is verticl line. It only hs n x intercept. Its slope is undefined. If the eqution solves to y = ny number, then the grph is horizontl line. It only hs y intercept. Its slope is 0 (zero). When grphing liner inequlity, the line will be dotted if the inequlity sign is < or >. If the inequlity signs re either or, the line on the grph will be solid line. Shde bove the line when the inequlity sign is or >. Shde below the line when the inequlity sign is < or. Inequlities of the form x >, x, x <, or x number, drw verticl line (solid or dotted). Shde to the right for > or. Shde to the left for < or. Remember: Dividing or multiplying by negtive number will reverse the direction of the inequlity sign.
3 x+ y = 6 3xy 6 y = 5 x+ 3 y 3 x x + 1< 3 x < 5 Grph the following: 1. x y = 4. x + 3y > x + y y 6
4 SKILL 1.3 Identify or interpret the slope nd intercepts of liner grph or liner eqution. To find the y intercept, substitute 0 for x nd solve for y. This is the y intercept. The y intercept is lso the vlue of b in y = mx + b. To find the x intercept, substitute 0 for y nd solve for x. This is the x intercept. 1. Find the slope nd intercepts of 3x + y = 14. 3x+ y = 14 y = 3x+ 14 y = 3 x+ 7 The slope of the line is 3, the vlue of m. The y intercept of the line is 7. The intercepts cn lso be found by substituting 0 in plce of the other vrible in the eqution. To find the y intercept: To find the x intercept: let x = 0; 3(0) + y = 14 let y = 0; 3 x + (0) = y = 14 3 x + 0 = 14 y = 14 3 x = 14 y = 7 x = 14 3 (0,7) is the y intercept. ( 14 3,0) is the x intercept. Find the slope nd the intercepts (if they exist) for these equtions: 1. 5x+ 7y = 70. x y = x + 3y = 3(5 + y) 4. x+ 5y = 15
5 SKILL 1.4 Determine the eqution of line, given the pproprite informtion such s two points, pointslope, slopeintercept, or its grph. The pointslope form for the eqution of line: ( y y ) = m( x x ) 1 1 m 1 1 where is the slope nd ( x, y ) is the point.  The eqution of grph cn be found by finding its slope nd its y intercept. To find the slope, find points on the grph where coordintes re integer vlues. Using points: ( x, y ) nd ( x, y ). 1 1 slope = y x y x 1 1 The y intercept is the y coordinte of the point where line crosses the y xis. The eqution cn be written in slopeintercept form, which is y = mx + b, where m is the slope nd b is the y intercept. To rewrite the eqution into some other form, multiply ech term by the common denomintor of ll the frctions. Then rerrnge terms s necessry.  Given two points on line, the first thing to do is to find the slope of the line. If points on the grph re ( x1, y1) nd ( x, y ), then the slope is found using the formul: y y1 slope = x x 1 The slope will now be denoted by the letter m. To write the eqution of line, choose either point. Substitute them into the formul: ( ) Y y = m X x Remember (, ) cn be (, ) or (, ) x y x y x y If m, the vlue of the 1 1 slope, is distributed through the prentheses, the eqution cn be rewritten into other forms of the eqution of line.
6 Find the eqution of line through (9, 6) nd ( 1,). x Cos63 = 1 Y y = m( X x ) Y = 4 5( X 1) Y = 4 5( X + 1) Y = 4 5X4 5 Y = 4 5 X This is the slopeintercept form. Multiplying by 5 to eliminte frctions, it is: 5Y = 4X + 6 4X + 5Y = 6 Stndrd form. Write the eqution of line through these two points: 1. (5,8) nd ( 3,). (11,10) nd (11, 3) 3. ( 4,6) nd (6,1) 4. (7,5) nd ( 3,5)
7 SKILL 1.5 Solve problems involving the use of equtions contining rtionl lgebric expressions. Add or subtrct rtionl lgebric frctions.  In order to dd or subtrct rtionl expressions, they must hve common denomintor. If they don't hve common denomintor, then fctor the denomintors to determine wht fctors re missing from ech denomintor to mke the LCD. Multiply both numertor nd denomintor by the missing fctor(s). Once the frctions hve common denomintor, dd or subtrct their numertors, but keep the common denomintor the sme. Fctor the numertor if possible nd reduce if there re ny fctors tht cn be cncelled. 1. Find the lest common denomintor for b nd 4 b. 3 3 These fctor into 3 b nd b. The first expression needs to be multiplied by nother nd b. The other expression needs to be multiplied by 3 nd. Then both expressions would be 3 3 b 3 = 1 3 b 3 = LCD.. Find the LCD for x 4, x + 5x+ 6, nd x + x 6. x 4 fctors into ( x)( x+ ) x + 5x+ 6 fctors into ( x+ 3)( x+ ) x + x 6 fctors into ( x+ 3)( x) To mke these lists of fctors the sme, they must ll be ( x + 3)( x + )( x ). This is the LCD ( b) 1(3 ) 10b 3 10b+ 3 + = + = + = b 4b 6 b( b) 4 b(3 ) 1b 1b 1b This will not reduce s ll 3 terms re not divisible by nything.
8 = x 4 x + 5x+ 6 x + x = ( x )( x + ) ( x+ 3)( x+ ) ( x+ 3)( x) ( x + 3) 3( x ) 7( x+ ) + = ( x )( x + )( x + 3) ( x+ 3)( x + )( x ) ( x+ 3)( x )( x+ ) x + 6 3x 6 7x = ( x )( x + )( x + 3) ( x+ 3)( x + )( x ) ( x+ 3)( x )( x+ ) x + 6 (3x 6) + 7x x + 6 = ( x + 3)( x )( x+ ) ( x+ 3)( x )( x+ ) Try These: This will not reduce x 3 x b 5 b x+ 3 x6 + x 5 x x15
9 Solve word problems with rtionl lgebric expressions nd equtions.  Some problems cn be solved using equtions with rtionl expressions. First write the eqution. To solve it, multiply ech term by the LCD of ll frctions. This will cncel out ll of the denomintors nd give n equivlent lgebric eqution tht cn be solved. 1. The denomintor of frction is two less thn three times the numertor. If 3 is dded to both the numertor nd denomintor, the new frction equls 1. x x + 3 originl frction: revised frction: 3x 3x + 1 x = x+ 6 = 3x+ 1 3x + 1 x = 5 5 originl frction: 13. Elly Me cn feed the nimls in 15 minutes. Jethro cn feed them in 10 minutes. How long will it tke them if they work together? Solution: If Elly Me cn feed the nimls in 15 minutes, then she could feed 115 of them in 1 minute, 15 of them in minutes, x 15 of them in x minutes. In the sme fshion Jethro could feed x 10 of them in x minutes. Together they complete 1 job. The eqution is: x x + = Multiply ech term by the LCD of 30: x+ 3x = 30 x = 6 minutes
10 3. A slesmn drove 480 miles from Pittsburgh to Hrtford. The next dy he returned the sme distnce to Pittsburgh in hlf n hour less time thn his originl trip took, becuse he incresed his verge speed by 4 mph. Find his originl speed. Since distnce = rte x time then time = distnce rte originl time 1 hour = shorter return time = 480 x x + 4 Multiplying by the LCD of xx ( + 4), the eqution becomes: ( x + ) x( x + ) = ( x) x x 4x = 960x x + 4x 3840 = 0 ( x+ 64)( x 60) = 0 Either (x60=0) or (x+64=0) or both=0 x = mph is the originl speed. x = 64 This is the solution since the time cnnot be negtive. Check your nswer = = = 7 Try these: 1. Working together, Lrry, Moe, nd Curly cn pint n elephnt in 3 minutes. Working lone, it would tke Lrry 10 minutes or Moe 6 minutes to pint the elephnt. How long would it tke Curly to pint the elephnt if he worked lone?. The denomintor of frction is 5 more thn twice the numertor. If the numertor is doubled, nd the denomintor is incresed by 5, the new frction is equl to 1. Find the originl number. 3. A trip from August, Mine to Glveston, Texs is 108 miles. If one cr drove 6 mph fster thn truck nd got to Glveston 3 hours before the truck, find the speeds of the cr nd truck.
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