algebra 2 help…solution sets….please?
3x – 8 = 4
soultion set {integers}
4x – 7 = 3
solution set {integers}
(2x – 6)(3x + 2) =0
soultion set {rational numbers}
!3x + 7! = 25
solution set {positive numbers}
x^2 = 49
solution set {real numbers}
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3x=12
x=4
solution set: {4}
4x-7=3 add 7 to both sides
4x=10 divide both sides by 5
x=10/4 simplify
x=5/2
solution set: {5/2}
(2x-6)(3x-2)=0 set each factor equal to zero
2x-6= 0 or 3x-2=0 add pure number to both sides
2x=6 or 3x=2 divide both sides by coefficient of x
x=3 or x=2/3
solution set: {3, 2/3}
!3x-7!=25
3x-7=25 or 3x-7=-25 add pure number to both sides
3x=32 or 3x=-18 divide both sides by 3
x=32/3 or x=-6
solution set: {32/3, -6}
x^2=36 take square root of both sides
x=7 or x=-7
solution set: {-7, 7}
x
3x = 8 + 4
3x = 12
x = 4
4x – 7 = 3
4x = 7 + 3
4x = 10
x = 10/4 or 2 1/2
(2x – 6) (3x + 2) = 0
2x – 6 = 0
2x = 6
x = 3
3x + 2 = 0
3x = -2
x = -2/3
!3x + 7! = 25
(3x + 7) = 25 or -(3x + 7) = 25
3x + 7 = 25 or -3x – 7 = 25
3x = 25 – 7 -3x = 25 + 7
3x = 18 -3x = 32
x = 6 or x = -32/3 or -10 2/3
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