Posadashannypeidn
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LINEAR EQUATION IN ONE VARIABLE

Sagot :

HuInDWorldidn
You're asking in two variables, yes? Then I'll give out a problem...

PROBLEM: 1) x - 2y = 3; 2) 3x + y = 1
SOLUTION: It depends upon you. You can use GRAPHING, where you would list the points which are the possible solutions to the problem - at least two points would do.
You could also use SUBSTITUTION, where you have to substitute a variable to solve the other variable.
You could also use ELIMINATION, where you could cross out a variable to solve the other one.
EQUATE:
GRAPHING:
              1) Points: (3,0) and (1, -1)
              2) Points: (0,1) and (1,0)
SUBSTITUTION:
                  [tex] \left \{ {{x - 2y = 3} \atop {3x + y = 1}} \right. \\ y = -3x + 1 \\ x -2(-3x + 1) = 3 \\ x +6x - 2 = 3 \\ 7x = 3 + 2 \\ 7x = 5 \\ x = \frac{5}{7} \\ \\ 3 ( \frac{5}{7}) + y = 1 \\ \frac{15}{7} + y = 1 \\ y = - \frac{15}{7} + 1 \\ y = - \frac{8}{7} [/tex]

ELIMINATION:
      [tex] \left \{ {x - 2y =3} \atop {3x + y =1}} \right. \\ 7x = 5 \\ x = \frac{5}{7} \\ \\ \frac{5}{7} - 2y = 3 \\ -2y = 3 - \frac{5}{7} \\ -2y = - \frac{16}{7} \\ y = \frac{7}{8} [/tex]
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