# ￼￼5(x1) +3(x2) +2(x3) =4 , 3(x1) +3(x2)+2(x3) =2 , x2 + x3 = 5 How to find x1,x2,x3 ?

##### 2 Answers

#### Explanation:

You have three equations with three unknowns, so all you have to do here is replace two of the unknowns into one equation and solve for the third one.

Your system of equations looks like this

#{(5x_1 + 3x_2 + 2x_3 = 4), (3x_1 + 3x_2 + 2x_3 = 2), (x_2 + x_3 = 5) :}#

Let' say that we want to solve for

#x_3 = 5 - x_2#

Plug this into the first two equations to get

#5x_1 + 3x_2 + 2 * (5 - x_2) = 4#

#5x_1 + 3x_2 + 10 - 2x_2 = 4#

#5x_1 + x_2 = - 6#

and

#3x_1 + 3x_2 + 2x_3 =2#

#3x_1 + 3 * x_2 + 2 * (5 - x_2) = 2#

#3x_1 + 3x_2 + 10 - 2x_2 = 2#

#3x_1 + x_2 = -8#

You have reduced the initial system of three equations with three unknowns to a system of two equations with two unknowns. Now multiply one of the two equations by

#{(5x_1 + x_2 = - 6 | * (-1)), (3x_1 + x_2 = -8) :}#

#{(-5x_1 - x_2 = 6), (3x_1 + x_2 = -8) :}#

Add these two equations by adding the left-hand sides and the right-hand sides separately

#-5x_1 - color(red)(cancel(color(black)(x_2))) + 3x_1 + color(red)(cancel(color(black)(x_2))) = -8 + 6#

#-2x_1 = -2 implies x_1 = ((-2))/((-2)) = color(green)(1)#

This means that

#3 * (1) + x_2 = -8#

#x_2 = -8 - 3 = color(green)(-11)#

Finally,

#x_3 = 5 - x_2 = 5 - (-11) = color(green)(16)#

**Alternatively**

You could also use **Cramer's Rule** to solve this system of equations, but I'm not sure if you've covered it in school already.

Here's how you can solve this system by using *Cramer's Rule*.

#### Explanation:

Start by writing down the system of equations

#{(5x_1 + 3x_2 + 2x_3 = 4), (3x_1 + 3x_2 + 2x_3 = 2), (x_2 + x_3 = 5) :}#

Now, to solve this system using **Cramer's Rule**, you need to write a *matrix* using the coefficients of the terms found on the left-hand sides of the equations and a *column* of the terms found on the right-hand sides of the equation

#|( 5, 3, 2), (3, 3, 2), (0, 1, 1) |" "# and#" "[(4), (2), (5) ]#

Next, find the **determinant**,

#D = 5 * |(3,2), (1,1)| - 3 * |(3,2), (0,1)| + 2 * |(3,3), (0,1) |#

#D = 5 * (3-2) - 3 * (3 - 0) + 2 * (3-0)#

#D = 5 - 9 + 6 = 2#

Next, you need to find the determinants of the coefficient matrix **with the answer column** in all three positions,

Start with

#|( color(red)(4), 3, 2), (color(red)(2), 3, 2), (color(red)(5), 1, 1) |" "#

#D_(x_1) = 4 * |(3,2), (1,1)| + 3 * |(2,2), (5,1)| + 2 * |(2,3), (5,1) |#

#D_(x_1) = 4 * (3-2) - 3 * (2 - 10) + 2 * (2-15)#

#D_(x_1) = 4 + 24 - 26 = 2#

Now do the same for

#|( 5, color(red)(4), 2), (3, color(red)(2), 2), (0, color(red)(5), 1) |" "#

#D_(x_2) = 5 * |(2,2), (5,1)| - 4 * |(3,2), (0,1)| + 2 * |(3,2), (0,5)|#

#D_(x_2) = 5 * (2-10) - 4 * (3-0) + 2 * (15-0)#

#D_(x_2) = -40 - 12 + 30 = -22#

and

#|( 5, 3, color(red)(4)), (3, 3, color(red)(2)), (0, 1, color(red)(5)) |" "#

#D_(x_3) = 5 * |(3,2), (1,5)| - 3 * |(3,2), (0,5)| + 4 * |(3,3), (0,1)|#

#D_(x_3) = 5 * (15 - 2) - 3 * (15 - 0) + 4 * (3 - 0)#

#D_(x_3) = 65 - 45 + 12 = 32#

Now, according to **Cramer's Rule**, you have

#x_1 = D_(x_1)/D" "# ,#" "x_2 = D_(x_2)/D" "# ,#" "x_3 = D_(x_3)/D#

This means that you have

#x_1 = 2/2 = color(green)(1)#

#x_2 = (-22)/2 = color(green)(-11)#

#x_3 = 32/2 = color(green)(16)#