Answers: ___________ 1) [B]: {-5, 5} 2) [A]: {-24, 8} 3) [C]: {-1, 4} 4) [A]: {-21, 15} 5) [C]: {-5/3, 3} 6) [A]: {⅔, 2} ____________ Explanation: _________________ 2) "Solve: | m + 8| = 16 "; Assuming we are to solve for "m"; we would use: ____________ "Case 1" and "Case 2" scenarios; _______________ since we need to find the value for "m" when: __________________ Case 1: (m + 8) = 16; AND: _____________ Case 2: - (m + 8) = 16 . _________________ → Let's solve, starting with Case 1: _____________________ m + 8 = 16; → Subtract "8" from EACH SIDE of the equation; _____________________ → m + 8 − 8 = 16 − 8 ; → To get: → m = 8 ______________ → Let us solve for "Case 2" : ____________ - (m + 8) = 16 ; __________________________ Note the distributive property of multiplication: _____________________ → a*(b + c) = ab + ac ; ___________________________ → As such: - (m + 8) = 16 ; ____________________________ Consider this equation as: _____________________ - 1(m + 8) = 16 ; _______________________ → Note the "-1" is implied, since anything multiplied by "1" is that same thing. The "negative" in the "negative 1" comes from the "negative sign" already present. _________________________ → - 1(m + 8) = 16 ; → (-1*m) + (-1*8) = 16 ;
→ -1m + (-8) = 16 ; → Rewrite as: -1m − 8 = 16 ; _______________ → Add "8" to EACH SIDE of the equation ; → -1m − 8 + 8 = 16 + 8 ; → to get: -1m = 24 ; ________________ → Divide EACH SIDE of the equation by "-1"; to isolate "m" on one side of the equation; & to solve for "m" ; → -1m / -1 = 24 / -1 ; → m = -24 _____________ So we have: m = 8; and m = -24 ; which is: "Answer choice: [A]: {-24, 8}" . _____________________ 3) Given: |2n - 3| = 5 ; Solve for "n" ; _______________ Case 1: 2n − 3 = 5 ; _______________ → Add "3" to EACH side of the equation; 2n − 3 + 3 = 5 + 3 ; → 2n = 8 ; → Now, divide EACH SIDE of the equation by "2"; to isolate "n" on one side of the equation; and to solve for "n"; → 2n / 2 = 8 / 2 ; → n = 4 . __________ Case 2: - (2n − 3) = 5 ; _________________ Simplify: - (2n − 3); → - 1(2n − 3) = (-1*2n) − (-1*3) = -2n − (-3) = -2n + 3 ; _________________ → Rewrite the equation, "- (2n − 3) = 5 " ; as: → -2n + 3 = 5; → Now, subtract "3" from EACH side of the equation: → -2n + 3 − 3 = 5 − 3 ; → -2n = 2 ; _____________ → Divide EACH SIDE of the equation by "-2"; to isolate "n" on one side of the equation; and to solve for "n" ; → -2n / -2 = 2 / -2 ; → n = -1 . → So, n = 4, AND -1; which is: Answer choice: [C]: {-1, 4} . __________________________ 4) Given: |x/3 + 1| = 6 ; Solve for "x"; ____________ Case 1: (x/3) + 1 = 6; Subtract "1" from EACH side of the equation; _______________ → (x/3) + 1 − 1 = 6 − 1 ; → (x/3) = 5 ; → x = 5*3 = 15 _____________ Case 2: - [ (x/3) + 1] = 6; → -1 [(x/3) + 1] = 6 ; → [-1*(x/3) ] + (-1*1) = 6 ; ______________ → - (x/3) + (-1) = 6 ; → - (x/3) − 1 = 6 ; → Add "1" to EACH side of the equation: → - (x/3) − 1 + 1 = 6 + 1 ; → - (x/3) − 1 + 1 = 6 + 1 ; → - (x/3) = 7 ↔ -x / 3 = 7; -1x = 7*3 ; → -1x = 21; ____________ → Divide EACH SIDE of the equation by "-1"; to isolate "x" on one side of the equation; & to solve for "x" ; ____________ → -1x / -1 = 21 / -1 ; → x = -21; So, x = 15, AND -21; which is: Answer choice: [A]: {-21, 15} . _____________ 5) Given: |2 − 3x| = 7 ; Solve for "x"; __________ Case 1: (2 − 3x) = 7 ______________ Given: 2 − 3x = 7 ; → Subtract "2" from EACH side of the equation; ______________ → 2 − 3x − 2 = 7 − 2 ; → to get: -3x = 5 ; ____________ → Now, divide EACH side of the equation by "-3"; to isolate "x" on one side of the equation; and to solve for "x" ; → -3x / =3 = 5/ -3 = - (5/3). ________ →Case 2: -(2 − 3x) = 7 ; Simplify, ________________________ Note: The distributive property of multiplication: → a*(b + c) = ab + ac ; AND: → a*(b − c) = ab − ac ; __________ → - (2 − 3x) = 7 ; → -1(2 − 3x) = 7 ; → (-1*2) − (-1*3x) = 7 ; _________ → - 2 − (-3x) = 7 ; → - 2 + 3x = 7 ; → Add "2" to EACH side of the equation ; → - 2 + 3x + 2 = 7 + 2 ; → 3x = 9 ; ____ → Now, divide EACH side of the equation by "3"; to isolate "x" on one side of the equation; and to solve for "x" ; → 3x / 3 = 9 / 3 ; x = 3; _____ So, x = - 5/3; AND 3; which is: Answer choice: [C]: {-5/3, 3} . ________________________ 6) Given: |2y − 2| = y ; Solve for "y"; ____________ Case 1: 2y − 2 = y; __________ → Subtract "y" from EACH SIDE of the equation; ___________ → 2y − 2 = y ; → 2y − 2 − y = y − y ; → y − 2 = 0 ; _____________ → Add "2" to EACH SIDE of the equation; to isolate "y" on one side of the equation; & to solve for "y" ; → y − 2 + 2 = 0 + 2 ; → y = 2 ________________ Case 2: - (2y − 2) = y _________________________ → Simplify: - (2y − 2); → -1(2y − 2) = (-1*2y) − (-1*2) = -2y - (-2) = -2y + 2 ;
→ Rewrite: - (2y − 2) = y ; as: → -2y + 2 = y ;
→ Add "2y" to EACH SIDE of the equation: → -2y + 2 + 2y = y + 2y ; → 2 = 3y ;
→ Now, divide EACH side of the equation by "3"; to isolate "y" on one side of the equation & to solve for "y" ; → 2 / 3 = 3y / 3 ; → ⅔ = y ↔ y = ⅔ ; so y = 2, AND ⅔; which is: → Answer choice: [A]: {⅔, 2} ________________
