find two numbers such that the sum of twice the first and thrice the second is 92 ,and four times the first exceeds seven times the second by 2

First number = 25

Second number  = 14

Step-by-step explanation:

Let the numbers be x & y

The sum of twice the first and thrice the second is 92

2x + 3y = 92   (I)

Four times the first exceeds seven times the second by 2

4x - 7y = 2  (II)

Multiply  equation (I) by -2 and then add the equations

(I)*-2          -4x - 6y = -184

(II)               4x - 7y = 2      { Now add, thus x will be eliminated}

                      - 13y = -182 {Divide both sides by -13}

y = -182/-13

y = 14

Plugin the value of y in equation (I)

2x + 3*14 = 92

2x + 42 = 92

       2x = 92 - 42

       2x = 50

        x  = 50/2

x = 25

First number = 25

Second number  = 14

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!