(5)
2r + s = 11 → (1)
r - s = 2 → (2)
add (1) and (2) term by term
3r = 13 ⇒ r = [tex]\frac{13}{3}[/tex]
substitute this value into (2)
- s = [tex]\frac{6}{3}[/tex] - [tex]\frac{13}{3}[/tex] = - [tex]\frac{7}{3}[/tex]
hence s = [tex]\frac{7}{3}[/tex]
(6)
substitute y = 3 - x into the other equation
5x + 3(3 - x) = - 1
5x + 9 - 3x = - 1
2x + 9 = - 1 ( subtract 9 from both sides )
2x = - 10 ( divide both sides by 2 )
x = - 5
substitute x = - 5 into y = 3 - x
y = 3 - (- 5) = 3 + 5 = 8
(7)
add both equations to eliminate the term in b
22a = 0 ⇒ a = 0 and b = 0
(8)
Since both equations express y in terms of x , equate the right sides
3x - 1 = 2x - 5 ( subtract 2x from both sides )
x - 1 = - 5 ( add 1 to both sides )
x = - 4
substitute this value into either of the 2 equations
y = - 8 - 5 = - 13
(9)
2y = 8 - 7x → (1)
4y = 16 - 14x → (2)
multiply (1) by 2
4y = 16 - 14x
Both equations are equal hence any value of x will make them true
example (1, [tex]\frac{1}{2}[/tex]) is a possible solution
The system has an infinite number of solutions
