

- <u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>



<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>


<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>





<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>





Hence, The value of x and y is 2 and 10/3
Given:
Joining fee = $28
Fee of each event = $4
To find:
Total cost for someone to attend 4 events.
Solution:
Let the number of events be x and total fee be y.
Fee for 1 event = $4
Fee for x events = $4x
Joining fee remains constant. So, the total fee is

Substitute x=4 in this equation.



Therefore, total cost of 4 events is $44.
Answer:
blue = 35
green = 29
red = 38
Step-by-step explanation:
Let r = red
Let g = green
let b = blue
r + g + b = 102
r = b + 3
b = g + 6 Subtract 6 from both sides of the equation
b - 6 = g
===================
substitute for red and green
(b + 3) + b + b - 6 = 102
b + 3 + b + b - 6 = 102
Combine like terms
3b - 3 = 102
Add 3 to both sides
3b - 3 + 3 = 102 + 3
3b = 105
Divide by 3
3b/3 = 105/3
b = 35
======================
r = b + 3
r = 35 + 3
r = 38
=======================
g = b - 6
g = 35 - 6
g = 29
========================
Answer:
Step-by-step explanation:
1,2,3,3,3,3,5,6,7,10,10,10,11,11,12,12,13,13,
14,15