Jeremy is 5 years younger than his older sister. His older sister is 9 years older than his younger sister. The total of their age is 49. Write and solve an equation to find the ages of Jeremy and his sisters. I need help please!

Answers

Answer 1
Answer: Jeremy=x
Older sister=x+5
younger sister=x+5-9 or x-4
x+(x+5)+(x-4)=49
combine like terms
3x+5-4=49
3x+1=49
subtract 1 from both sides
3x=48
divide both sides by 3
x=16
Jeremy=16
Older sister=16+5=21
younger sister =16-4 or 16+5-9=12
J=16
OS=21
YS=12
Answer 2
Answer: Jeremy - x
Older sister - x+5
Younger sister - x-4
3x+1=49
3x=48
X= 16

Jeremy: 16
Older sister: 21
Younger sister: 12

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The linear combination method is applied to a system of equations as shown.4(.25x + .5y = 3.75) → x + 2y = 15
(4x – 8y = 12) → x – 2y = 3
2x = 18
What is the solution of the system of equations?

(1,2)
(3,9)
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Answers

2x = 18
x = 18/2
x = 9

4(.25x + .5y = 3.75) → x + 2y = 15
(4x – 8y = 12) → x – 2y = 3

- 2y = 3 - x
- 2y = 3 - 9
- 2y = - 6
y = - 6/ - 2
y = 3

(9,3)


Answer:

D

Step-by-step explanation:

(9,3)

just did it

$6300 is invested , part of it at 12%, and part at 6%.  For a certain year, the total yield is 564.00.  How much was invested at each rate?

Answers

x - first part of the investment (12% rate)
y - second part of the investment (6% rate)

Total investment: x + y = 6300
Total yield: 564 = x*0,12 + y*0,06

Based on that we can calculate x and y:
\left \{ {y=6300-x} \atop {0,12x+0,06(6300-x)=564}} \right.

\left \{ {y=6300-x} \atop {0,12x+378-0,06x=564}} \right.

\left \{ {y=6300-x} \atop {0,06x=186}} \right.

\left \{ {y=6300-x} \atop {x=3100}} \right.

\left \{ {y=3200} \atop {x=3100}} \right.

Answer: $3100 was invested at 12% and $3200 at 6%.

If you have any questions, please let me know!

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Answers

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What is the sum of the quotients of 25x^2/5x and the quotient of 8x^2/x

Answers

If you would like to solve (25 * x^2) / (5 * x) + (8 * x^2) / x, you can calculate this using the following steps:

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The correct result would be 13 * x.

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Answers


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Answers

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Answer

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