SOLUTION: Please help me solve for I for this equation: {{{ 3I1-4I2=17 and I1+3I2=-3 }}}. Note the I1 (has a number 1 at the bottom right hand corner and the I2 has the number 2 on the ri

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me solve for I for this equation: {{{ 3I1-4I2=17 and I1+3I2=-3 }}}. Note the I1 (has a number 1 at the bottom right hand corner and the I2 has the number 2 on the ri      Log On


   

Question 68645: Please help me solve for I for this equation: +3I1-4I2=17+and+I1%2B3I2=-3+.
Note the I1 (has a number 1 at the bottom right hand corner and the I2 has the number 2 on the right hand corner.)
Answer by rmromero(383) About Me  (Show Source):
You can put this solution on YOUR website!
+3I%5B1%5D-4I%5B2%5D=17+and+I%5B1%5D%2B3I%5B2%5D=-3+


Let us represent other variables.
x+=+I%5B1%5D
y+=+I%5B2%5D


Rewrite the equations
3x-4y=17
x+3y=-3


We can solve for the variables using elimination method.
Multiply -3 to x+3y=-3
3x - 4y = 17
-3x - 9y = 9


Add the two equations
-13y = 26

Divide -13 both sides
y = -2 <<<<<<<<<<<<<<<y+=+I%5B2%5D


Substitute 2 for y in either equation. Solve for x.
x+3y=-3, y = 2
x + 3(-2)= -3
x - 6 = -3
x = 3 <<<<<<<<<<<<<< x+=+I%5B1%5D

Checking:
3x-4y=17 , x = 3, y = -2
3(3) - 4(-2) = 17
9 + 8 = 17
17 = 17 ----------->> True


x+3y=-3, x = 3, y = -2
3 + 3(-2) = -3
3 -6 = -3
-3 = -3 --------->> True