Substitute each value of x in y=2x-3
so,
first box (-2)
y= 2(-2)-3
y= -4-3
y= -7
second box (0)
y= 2(0)-3
y= 0-3
y= -3
third box (3)
y=2(3)-3
y=6-3
y=3
Answer:
Terms must have the same variable (letter) and the same exponent (little number)
(7x² +3y+ 5) +(9x²+11y- 2)
Opening bracket
7x²+3y+5+9x²+11y-2
keeping like terms together
7x²+9x²+3y+11y+5-2
Since terms having same variable and exponent can be subtracted, added,divided and multiplied
So
Solving like terms we get
<u>16x²+14y+3</u> which is a correct answer.
ANSWER
(-2,-1)
(3,14)
EXPLANATION
The given system is


or

Equate both of them:

This implies that,


x=-2 or x=3
When x=-2, y=y=3(-2)+5=-1
(-2,-1) is a solution.
When x=3 , y=3(3)+5=14
(3,14) is also a solution.