Assume the smaller one is x, then the other ones will be x+2 and x+4
x(x+2) + 2 = 5(x+4)
x^2 + 2x - 5x - 18 = 0
-> x^2 - 3x - 18 = 0
-> (x-6)(x+3) = 0
-> x = 6 or -3, since x must be positive, then x = 6
So the numbers are: 6, 8, 10
Answer:
(2, 4)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = 2x
x = -y + 6
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: y = 2(-y + 6)
- Distribute 2: y = -2y + 12
- Isolate <em>y</em> terms: 3y = 12
- Isolate <em>y</em>: y = 4
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define equation: x = -y + 6
- Substitute in <em>y</em>: x = -4 + 6
- Add: x = 2
We use the formula a^2 - b^2 = ( a - b )( a + b );
We have a = 5m - 2 and b = 3m - 4;
<span>(5m-2)^2-(3m-4)^2 = (5m - 2 -3m + 4) x (5m-2 + 3m - 4) = (2m + 2)(8m - 6) = 2(m +1) x 2(4m - 3) = 4(m+1)(4m-3);
</span>
The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t
x=10 m/st ⇒ x=10t (1)
Distance traveled by car B=x
time of car B=t-3
x=15(t-3) (2)
With the equations (1) and (2) we make a system of equations:
x=10t
x=15(t-3)
We solve this system of equations:
10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9
t-3=9-3=6
x=10 t=10 (9)=90
Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.
