Acontainer holds a combination of dimes and quarters. there are 178 coins in all, and their total value is $29.80. the system of equations models this situation, where x is the number of dimes and y is the number of quarters.
x+y=178
0.1x+0.25y=29.8
how many dimes and quarters are in the container?

a.
123 dimes and 55 quarters

b.
114 dimes and 64 quarters

c.
102 dimes and 76 quarters

d.
98 dimes and 80 quarters

x+y=178

0.1x+0.25y=29.8

Solve x+y =178 for x

Add -y to both sides

x+y+-y=178+-y

x= -y+178

Substitute -y+178 for x in 0.1x+0.25y=29.8

0.1x+0.25y=29.8

0.1(-y+178)+0.25y=29.8

Simplify both sides of the equation

0.15y+17.8=29.8

Add -17.8 to both sides

0.15y+17.8+-17.8=29.8+-17.8

0.15y=12

Divide both sides by 0.15

0.15y/0.15= 12/0.15

y=80

Substitute 80 for y in x = -y+178

x= -y+178

x= -80+178

Simplify both sides of the equation

x= 98

D

x=98 and y= 80

I hope that's help !

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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