Chapte 1 Section : Fomulas nswes to Poblems Solve the following equations fo the vaiable indicated. 1.* Solve fo x: 5x 10. Solve fo e: 7ae 3u x 3u e 7a 3.* Solve fo a: is at 4. Solve fo v: 4v is a t v 5.* Solve fo n: w 3nu 6. Solve fo B: B 3 w 3 n B 3u 7.* Solve fo T: T 5 T 3S 8. Solve fo D: 8D 5 5D T S D 5 3 9.* Solve fo : 3 Z M 10. Solve fo L: al + hl 9 3 9 L Z M a + h 11.* Solve fo : V 4B 1. Solve fo : π h V 4B π h 13.* Solve fo Q: Qx 3Wi 4Q 14. Solve fo t: mg Wt + Mt Q 3Wi x + 4 mg t W + M 15.* Solve fo m: mc 16. Solve fo m: m c mv m v *See xplanations & Woked Solutions on page 4. Page 1 of 13
Chapte 1 Section : Fomulas 17.* Solve fo : BC DF 18.* Solve fo : BC DF D F BC BCF D 19.* Solve fo Z: C BD Z ZF 0. Solve fo h: C g 3aQ 3hQ g h 9aQ C BDF 1.* Solve fo : 4 + 3B B 7B. Solve fo C: C 5B B 4C B 5B B C B 3.* Solve fo : 3x 5Bx 7x 4. Solve fo x: 3x 5Bx 7x 5Bx 7x 3x x 3 5B+ 7 5.* Solve fo B: 3x 5Bx 7x 6.* Solve fo S: B 3x + 7x 5x 5 3 R S T 5RT S T 3R 7.* Solve fo T: 9.* Solve fo x: 5 3 R S T 8. Solve fo : 3RS T S 5R 1 1 1 30. Solve fo W: x x x + 1 1 1 1 7 x 7x x 1 7 3 + W W W 3 31.* Solve fo h: 3π ( h+ ) 3. Solve fo : B 3B( B 3) h 3π 6π 11 *See xplanations & Woked Solutions on page 4. Page of 13
Chapte 1 Section : Fomulas 33.* Solve fo : B( x ) C( x B) 34. Solve fo x: B( x ) C( x B) C( x B) B ( C) Bx ( ) x B C 35.* Solve fo B: B( x ) C( x B) 36. Solve fo x: Cx B x + C 1 3 + 1 x x 1 x 3 37.* Solve fo : 39.* Solve fo B: 5x + 1 1 x 38. Solve fo : x x x + 5x+ 1 1 + 6 3B 40. Solve fo z: 5 B 1 + 6 3B 5 B x x + z x x z x + *See xplanations & Woked Solutions on page 4. Page 3 of 13
Chapte 1 Section : Fomulas *xplanations & Woked Solutions 1. Solve fo x: 5x 10 3. Solve fo a: is at 5x 10 is at 5x 10 is a t 5 5 t t 5 x 5 is a 5 5 t x 5. Solve fo n: w 3nu 7. Solve fo T: T 5T 3S w 3nu T 5T 3S w 3 n u 5T 5T 3u 3u 3T 3S w n 3u 3 T 3 S 3 3 T S 9. Solve fo : 3 Z M 11. Solve fo : V 4B Stat b factoing out an : V 4B 3 Z M 3 ( Z M) V 4B 4B 4B 3 ( Z M ) V Z M ( Z M) 4B 3 Z M 13. Solve fo Q: Qx 3Wi 4Q Stat b collecting all Q tems on the left side of the equation: Qx 3Wi 4Q + 4Q + 4Q Qx + 4Q 3Wi Now facto out a Q: Qx + 4Q 3Wi Q x + 4 3Wi Page 4 of 13
Chapte 1 Section : Fomulas Finall, divide both sides of the equation b the facto ( x + 4) : Q( x + 4) 3Wi ( x + 4) ( x + 4) Q 3Wi x + 4 15. Solve fo m: mc c c mc m c mc 17. Solve fo : BC DF BC DF BC DF DF D F BC DF 18. Solve fo : D BC F Fist, eliminate the factions. This is accomplished b multipling both sides of the equation times the LCM (least common multiple) of all the denominatos in the equation. Fo this poblem, the LCM is BC. D F BC BC BC BC D BC F BC D BCF To get all b itself (thus solving the equation fo ), simpl divide both sides b D: D BCF D D D BCF D D BCF D 19. Solve fo Z: C BD ZF C Fist, eliminate the factions. This is accomplished b multipling both sides of the equation times the LCM (least common multiple) of all the denominatos in the equation. Fo this poblem, the LCM is BDC. Page 5 of 13
Chapte 1 Section : Fomulas C ZF BDC BDC BD C BDCC BD CZF BD C C BDFZ To get Z all b itself (thus solving the equation fo Z), divide both sides b BDF: C ZF BDC BDC BD C BDCC BD CZF BD C C BDFZ BDF BDF C BDF Z BDF BDF Z C BDF 1. Solve fo : 4 + 3B B 7B Get all the tems with on one side of the equal sign, with all othe tems on the opposite side (since we e solving fo ). Subtacting B fom both sides moves all the tems to the left side of the equal sign, and subtacting 3B fom both sides moves all tems lacking to the ight side: 4 + 3B B 7B B 3B B 3B 4 B 10B Facto out an on the left side of the equal sign, and then divide both sides of the equal sign b whateve is then multiplied times that : 4 B 10B ( 4 B) 10B ( 4 B) 10B ( 4 B) ( 4 B) ( 4 B) 10B ( 4 B) 4 B 10B 4 B Page 6 of 13
Chapte 1 Section : Fomulas Next, facto a out of the numeato and denominato, and cancel this common facto: ( B) ( B) 5 5B B 5B B This is the solution, but it can be simplified futhe b multipling the numeato and denominato b ( 1): 3. Solve fo : 3x 5Bx 7x ( B) ( B) 1 5 5B 1 B Get all the tems with on one side of the equal sign, with all othe tems on the opposite side (since we e solving fo ). Subtacting fom both sides moves all the tems to the left side of the equal sign, and adding 5Bx to both sides moves all tems lacking to the ight side: 3x 5Bx 7x + 5Bx + 5Bx 3x 5Bx 7x Facto out a on the left side of the equal sign, and then divide both sides of the equal sign b whateve is then multiplied times that : 3x 5Bx 7x 3 5Bx 7x ( x ) 5. Solve fo B: 3x 5Bx 7x ( 3x ) 5Bx 7 ( 3x ) ( 3x ) x ( 5B 7 ) 5Bx 7x o x 3x 3x Get all the tems with B on one side of the equal sign, with all othe tems on the opposite side (since we e solving fo B). Subtacting 3x to both sides accomplishes this: 3x 5Bx 7x 3x 3x 5Bx 7x 3x Page 7 of 13
Chapte 1 Section : Fomulas Divide both sides b 5x 5Bx 7x 3x 5x 5x 5Bx 5 x B 7x 3x 5x 7x 3x 5x This is the solution, but it can be simplified futhe b multipling the numeato and denominato b ( 1): 1 ( 7x 3x) B 1 5x 3x + 7x B 5x 6. Solve fo S: 5 3 R S T The easiest wa to deal with such poblems is to fist get id of all the factions. This is accomplished b multipling both sides of the equation times the LCM of all the denominatos in the equation. Fo this poblem, the LCM is RST, since all the denominatos divide into RST, and thee is nothing smalle that does the job. Multipling both sides of the equation times a quantit is the same as multipling each tem in the equation b that quantit (due to the distibutive popet): 5 3 RST RST R S T R ST RST 5 RS T 3 R S T ST 5RT 3RS Next, get all the tems with S on one side of the equal sign, and all othe tems on the opposite side of the equal sign (since we e solving fo S). This is accomplished b subtacting ST fom both sides of the equation: ST 5RT 3RS ST ST 5RT 3RS ST Finall, facto out the S, and divide both sides of the equation b what s (then) multiplied times the S: 5RT 3RS ST 5RT S 3R T Page 8 of 13
Chapte 1 Section : Fomulas 5RT S( 3R T) ( 3R T) ( 3R T ) 5RT S( 3R T) 3R T ( 3R T ) 5RT S 3R T This is the solution, but it can be simplified futhe b multipling the numeato and denominato b ( 1): 1 ( 5RT ) 5RT S 1 3R T 3R + T 5RT S T 3R 7. Solve fo T: 5 3 R S T The easiest wa to deal with such poblems is to fist get id of all the factions. This is accomplished b multipling both sides of the equation times the LCM of all the denominatos in the equation. Fo this poblem, the LCM is RST, since all the denominatos divide into RST, and thee is nothing smalle that does the job. Multipling both sides of the equation times a quantit is the same as multipling each tem in the equation b that quantit (due to the distibutive popet): 5 3 RST RST R S T R ST RST 5 RS T 3 R S T ST 5RT 3RS ll the tems with T ae on the left side of the equation, and the ight side of the equation has the onl tem lacking T. Facto out the T, and divide both sides of the equation b what s (then) multiplied times the T: ST 5RT 3RS ( S 5R) T 3RS ( S 5R) T 3RS ( S 5R) ( S 5R) ( S 5R) T 3RS ( S 5R) S 5R T 3RS S 5R Page 9 of 13
Chapte 1 Section : Fomulas 9. Solve fo x: 1 1 1 x x The easiest wa to deal with such poblems is to fist get id of all the factions. This is accomplished b multipling both sides of the equation times the LCM of all the denominatos in the equation. Fo this poblem, the LCM is x, since all the denominatos divide into x, and thee is nothing smalle that does the job. Multipling both sides of the equation times a quantit is the same as multipling each tem in the equation b that quantit (due to the distibutive popet): 1 1 1 x x x x x 1 x 1 x 1 x x x 1 The onl tem with x is on the left side of the equation. dding to both sides isolates the x tem: x 1 + + x 1 + Dividing both sides of the equation b then completes the pocess of solving fo x: 31. Solve fo h: 3π ( h+ ) Stat b distibuting the 3π x 1+ x 1+ x 1+, to get the h out of the paentheses: ( ) 3π h+ 3π h+ 3π 6πh + 3π Next, get all the tems with h on one side of the equal sign, and all othe tems on the opposite side of the equal sign (since we e solving fo h). This is accomplished b subtacting 3π fom both sides of the equation: Page 10 of 13
Chapte 1 Section : Fomulas Finall, divide both sides b 6π : 3π 6π h + 3π 3π 3π 6πh 3π 6πh 6π 6π 3π 6π 3π 6π 6π h 6 π h 33. Solve fo : B( x ) C( x B) This can be accomplished in one step! Simpl divide both sides of the equation b B(x ): Bx ( ) Bx ( ) B ( x ) B ( x ) ( ) C x B B( x ) ( ) C x B Bx ( ) ( ) C x B Bx ( ) 35. Solve fo B: B( x ) C( x B) Distibute on both sides of the equation, to eliminate the paentheses: B( x ) C x B Bx B Cx CB dd CB to both sides, to collect all the B tems on the left side of the equation: Bx B Cx CB + CB + CB Bx B + CB Cx Facto out the B, and then divide both sides b B s facto: Bx B + CB Cx ( x C) + B Cx Page 11 of 13
Chapte 1 Section : Fomulas ( x + C) B Cx ( x + C) ( x + C) ( x + C) B Cx ( x + C) x + C Cx B x + C 37. Solve fo : 5x + 1 1 x x x The easiest wa to deal with such poblems is to fist get id of all the factions. This is accomplished b multipling both sides of the equation times the LCM of all the denominatos in the equation. Fo this poblem, the LCM is x, since all the denominatos divide into x, and thee is nothing smalle that does the job. Multipling both sides of the equation times a quantit is the same as multipling each tem in the equation b that quantit (due to the distibutive popet). 5x + 1 1 x x x x x x ( 5x + 1) x 1 x x x x x ( 5x 1) + x x x 5x + 1 x Next, get all the tems with on one side of the equal sign, and all tems without on the othe side of the equal sign (because we e solving fo ). Fo this poblem all we need do is add x to both sides of the equation. 5x + 1 x + x x + 5x + 1 + x x 39. 1 Solve fo B: + 6 3B The easiest wa to deal with such poblems is to fist get id of all the factions. This is accomplished b multipling both sides of the equation times the LCM of all the denominatos in the equation. Fo this poblem, the LCM is, since all the denominatos divide into, and thee is nothing smalle that does the job. Multipling both sides of the equation times a quantit is the same as multipling each tem in the equation b that quantit (due to the distibutive popet). Page 1 of 13
Chapte 1 Section : Fomulas 1 + 6 3B ( 1) B + 6 6 3B 6 B 3B ( 1) + 6 3B B ( 1) + B 4 + B 5 Finall, divide both sides b : B 5 5 B Page 13 of 13