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October 1, 2014

October 1, 2014

Posted by **Darryl** on Monday, January 17, 2011 at 2:08pm.

Graph:

Substitution:

Elimination:

Show Your work!

(I can'tfigure out how to graph it and when i used substitution and elimination it came out with two different anwsers. I need help! and i need to be able to show my work!)

- Math-Algebra -
**Henry**, Tuesday, January 18, 2011 at 11:04pmEq1: 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.

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