The sum of three numbers is 131. the second of three numbers is seven more than twice the first. the third number is 12 less than the first. what equation can be used to solve the problem? x + x + x = 131 x + (x + 7) + (x – 12) = 131 x + (2x + 7) + (x – 12) = 131 x + (2x + 7) – 12 = 131

For value of x

x + (2x + 7) + (x - 12 ) = 131

For y

y = 2x + 7

For z

z = x -12

Step-by-step explanation:

Let three numbers be

x , y and z

Now according to first condition

x + y +z = 131 (i)

According to other condition

The second number i.e. y is 7 more then twice of first

which will be

y = 2x + 7             (ii)

According to last condition

3rd number i.e. z is 12 less then first which is

z = x - 12                (iii)

As we have got The value of y and z from equation (ii) and (iii) respectively we can find the value of x by substituting these values in equation (i)

Now substituting values of x and y in equation (i) gives us

x + (2x + 7) + (x - 12 ) = 131

From this equation we can find the value of x and which could be substituted to the equation (ii) to find the value of y and for value of z we can substitute it to equation (iii)

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.