Math, asked by khandelwaljitu1234, 21 days ago

show the points A(1,2) B(-1 ,-16) C(0,-7) lie on the graph of the linear equation y= 9x - 7​

Answers

Answered by Anjula
4

Answer:

Step-by-step explanation:

To show that the points lie on the graph of linear equation y=9x-7

We have to check for the given values

For 1st value/ordered pair ,

A(1,2) x = 1, y = 2

Substituting in the given eqn,

2=9(1)-7

=> 2=2

This ordered pair is satisfied

Second pair , B(-1,-16) x = -1,y=-16

=> -16 = 9(-1)-7

=> -16= -16

This pair also satisfies the eqn

Third pair , C(0,-7) x=0,y=-7

=> -7= 9(0)-7

=> -7 = -7

This one also satisfies the equation

So , all of the Values/pairs satisfies and are solutions of y=9x-7.

Hence they lie on the graph of linear equation y=9x-7

Answered by shervinsalaah
0

Before answering keep in mind,

*First number represented by the coordinate belongs to x axis (horizontal) and Second number represented by the coordinate belongs to y axis (vertical)

* A graph's equation y must be the subject.

*To show the points lie on the graph you can just substitute the values of x or y and see whether the other is coming as the answer.

*You have to follow BODMAS.

* While multiplying if both numbers have different signs, the answer will be negative; if they have same signs, answer will be positive.

Answer:

A = x coordinate is 1 and y coordinate is 2

y= 9x - 7​

 =9 x 1  - 7

 =9 - 7

  =2

Hence, when x=1, y=2.

So, this point lie on the graph of the linear equation.

B = x coordinate is -1 and y coordinate is -16

y= 9x - 7​

 =9 x -1  - 7

 = -9 - 7

  = -16

Hence, when x= -1, y= -16.

So, this point lie on the graph of the linear equation.

C = x coordinate is 0 and y coordinate is -7

y= 9x - 7​

 =9 x 0  - 7

 =0 - 7

  = -7

Hence, when x=0, y= -7.

So, this point lie on the graph of the linear equation.

As all the values satisfy the equation, it is shown that the points A(1,2) B(-1 ,-16) C(0,-7) lie on the graph of the linear equation.

If my answer was helpful, please mark me as brainliest.

Similar questions