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Fed [463]
3 years ago
12

Four times a number added to 3 times a larger number is 31. Seven subtracted from the larger number is equal to twice the smalle

r number. Let x represent the smaller number and y represent the larger number. Which equations represent this situation?
Mathematics
2 answers:
aev [14]3 years ago
5 0
Let the smaller number be x.
Let the bigger number be y.

\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y- 7 = 2x \tex{ ----- (2) } \end{cases}

Rearrange equation (2):
\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y = 2x + 7 \tex{ ----- (2) } \end{cases}



Sub (2) into (1):

4x + 3(2x + 7 ) = 31
4x + 6x + 21 = 31
10x + 21 = 31
10x = 10
x = 1

Sub x = 1 into (2):

y = 2x + 7
y = 2(1) + 7
y= 9

Answer: The two numbers are 1 and 9.

Gelneren [198K]3 years ago
3 0
4x+3y=31

y-7=2x

(y-7)/2=x
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emmainna [20.7K]

Answer:

I found two different solutions. Hope one of them help!

1. x = -1/3 = -0.333

2. x = 5/2 = 2.500

Step-by-step explanation:

13 ± √ 289

  x  =    ——————

                     12

Can  √ 289 be simplified ?

Yes!   The prime factorization of  289   is

  17•17

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 289   =  √ 17•17   =

               ±  17 • √ 1   =

               ±  17

So now we are looking at:

          x  =  ( 13 ± 17) / 12

Two real solutions:

x =(13+√289)/12=(13+17)/12= 2.500

or

x =(13-√289)/12=(13-17)/12= -0.333

Two solutions were found :

x = -1/3 = -0.333

x = 5/2 = 2.500

7 0
3 years ago
What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to
Assoli18 [71]

Answer:

x=  −41 /20

Step-by-step explanation:

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(4x+16x)+(6+24+11)=0(Combine Like Terms)

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20x+41=0

Step 2: Subtract 41 from both sides.

20x+41−41=0−41

20x=−41

Step 3: Divide both sides by 20.

20x /20  =  −41 /20

x= −41 /20

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