Posted by Darryl on Monday, January 17, 2011 at 2:08pm.
Eq1: 3X - 2Y = 10.
Eq2: 2X + 3Y = -2.
Substitution Method:
In Eq2, solve for X:
Eq2: 2X + 3Y = -2,
2X = -3Y - 2,
X = -3Y/2 - 1,
In Eq1, substitute -3Y/2 - 1 for X:
3(-3X/2 - 1) - 2Y = 10,
-9X/2 - 3 - 2Y = 10,
Multiply both sides by 2 and get:
-9X - 6 -4Y = 20,
-13Y = 20 + 6,
-13Y = 26,
Y = 26 / -13 = - 2.
In Eq2, substitute -2 for Y:
2X + 3*-2 = -2,
2X - 6 = - 2,
2X = -2 + 6,
2X = 4,
X = 4 / 2 = 2.
Solution Set = (2, -2).
Elimination Method:
Multiply Eq1 by -2 and Eq2 by 3 and get:
-6X + 4Y = -20,
+6X + 9Y = -6
Add the 2 Eqs and get:
13Y = -26,
Y = -26 / 13 = -2.
Substitute -2 for Y in Eq1:
3X - 2*-2 = 10,
3X + 4 = 10,
3X = 10 - 4,
3X = 6,
X = 6 / 3 = 2.
Solution Set = (2, -2).
GRAPH: Use the following points.
Eq1:(-2, -8) (0, -5), (2, -2).
Eq2: (-2.5, 1),(-4, 2) (2, -2).
The point where the lines intersect is the solution.