Question

Three times a number increased by five times another number is 49, but twice the second number exceeds five times the first number by 1. What are the numbers?​

Answer 1

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Let x = "a number"

Let y = "another number"

:

Write an equation for each statement:

:

"three times a number increased 5 times another number is 49"

3x + 5y = 49

:

"twice the second number exceeds 5 times the first number by 1."

2y = 5x + 1

In standard form

-5x + 2y = 1

:

what are the numbers?

:

Multiply the 1st equation by 5, and the 2nd equation by 3

15x + 25y = 245

-15x + 6y = 3

-----------------Adding eliminates, x; find y:

0x + 31y = 248

$y = \frac{248}{31}$

y = 8

:

Find x using the 1st equation

3x + 5(8) = 49

3x = 49 - 40

$x = \frac{9}{3}$

x = 3

Check solutions in the 2nd equation:

2(8) = 5(3) + 1

16 = 15 + 1; confirms our solution

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Answer 2

Answer:

3   and   8

Explanation:

3a + 5b = 49

2b = 5a + 1

then:

2b - 5a = 1       equation 1

-5a + 2b = 1     equation 2

then:

multiplyng by -2.5 the equation 2 and sum to the equation 1:

   3a + 5b = 49

12.5a  - 5b = -2,5

15.5a  + 0   = 46.5

15.5a = 46.5

a = 46.5/15.5

a = 3

2b = 5a + 1

2b = (5*3) + 1

2b = 15 + 1

2b = 16

b = 16/2

b = 8

Check:

3a + 5b = 49

3*3 + 5*8 = 49

9 + 40 = 49