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

Sep 10, 2015

$\left\{\begin{matrix}{x}_{1} = 1 \\ {x}_{2} = - 11 \\ {x}_{3} = 16\end{matrix}\right.$

#### 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

$\left\{\begin{matrix}5 {x}_{1} + 3 {x}_{2} + 2 {x}_{3} = 4 \\ 3 {x}_{1} + 3 {x}_{2} + 2 {x}_{3} = 2 \\ {x}_{2} + {x}_{3} = 5\end{matrix}\right.$

Let' say that we want to solve for ${x}_{1}$ first. Use the third equation to write ${x}_{3}$ as a function of ${x}_{2}$

${x}_{3} = 5 - {x}_{2}$

Plug this into the first two equations to get

$5 {x}_{1} + 3 {x}_{2} + 2 \cdot \left(5 - {x}_{2}\right) = 4$

$5 {x}_{1} + 3 {x}_{2} + 10 - 2 {x}_{2} = 4$

$5 {x}_{1} + {x}_{2} = - 6$

and

$3 {x}_{1} + 3 {x}_{2} + 2 {x}_{3} = 2$

$3 {x}_{1} + 3 \cdot {x}_{2} + 2 \cdot \left(5 - {x}_{2}\right) = 2$

$3 {x}_{1} + 3 {x}_{2} + 10 - 2 {x}_{2} = 2$

$3 {x}_{1} + {x}_{2} = - 8$

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 $\left(- 1\right)$ to get

$\left\{\begin{matrix}5 {x}_{1} + {x}_{2} = - 6 | \cdot \left(- 1\right) \\ 3 {x}_{1} + {x}_{2} = - 8\end{matrix}\right.$

$\left\{\begin{matrix}- 5 {x}_{1} - {x}_{2} = 6 \\ 3 {x}_{1} + {x}_{2} = - 8\end{matrix}\right.$

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

$- 5 {x}_{1} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}_{2}}}} + 3 {x}_{1} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}_{2}}}} = - 8 + 6$

$- 2 {x}_{1} = - 2 \implies {x}_{1} = \frac{\left(- 2\right)}{\left(- 2\right)} = \textcolor{g r e e n}{1}$

This means that ${x}_{2}$ is equal to

$3 \cdot \left(1\right) + {x}_{2} = - 8$

${x}_{2} = - 8 - 3 = \textcolor{g r e e n}{- 11}$

Finally, ${x}_{3}$ is equal to

${x}_{3} = 5 - {x}_{2} = 5 - \left(- 11\right) = \textcolor{g r e e n}{16}$

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.

Sep 10, 2015

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

#### Explanation:

Start by writing down the system of equations

$\left\{\begin{matrix}5 {x}_{1} + 3 {x}_{2} + 2 {x}_{3} = 4 \\ 3 {x}_{1} + 3 {x}_{2} + 2 {x}_{3} = 2 \\ {x}_{2} + {x}_{3} = 5\end{matrix}\right.$

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

$| \left(5 , 3 , 2\right) , \left(3 , 3 , 2\right) , \left(0 , 1 , 1\right) | \text{ }$ and $\text{ } \left[\begin{matrix}4 \\ 2 \\ 5\end{matrix}\right]$

Next, find the determinant, $D$, of the coefficient matrix

$D = 5 \cdot | \left(3 , 2\right) , \left(1 , 1\right) | - 3 \cdot | \left(3 , 2\right) , \left(0 , 1\right) | + 2 \cdot | \left(3 , 3\right) , \left(0 , 1\right) |$

$D = 5 \cdot \left(3 - 2\right) - 3 \cdot \left(3 - 0\right) + 2 \cdot \left(3 - 0\right)$

$D = 5 - 9 + 6 = 2$

Next, you need to find the determinants of the coefficient matrix with the answer column in all three positions, ${x}_{1}$, ${x}_{2}$, and ${x}_{3}$.

Start with ${D}_{{x}_{1}}$, which will be calculated for a matrix that has the $x$-column replaced with the answers-column

$| \left(\textcolor{red}{4} , 3 , 2\right) , \left(\textcolor{red}{2} , 3 , 2\right) , \left(\textcolor{red}{5} , 1 , 1\right) | \text{ }$

${D}_{{x}_{1}} = 4 \cdot | \left(3 , 2\right) , \left(1 , 1\right) | + 3 \cdot | \left(2 , 2\right) , \left(5 , 1\right) | + 2 \cdot | \left(2 , 3\right) , \left(5 , 1\right) |$

${D}_{{x}_{1}} = 4 \cdot \left(3 - 2\right) - 3 \cdot \left(2 - 10\right) + 2 \cdot \left(2 - 15\right)$

${D}_{{x}_{1}} = 4 + 24 - 26 = 2$

Now do the same for ${D}_{{x}_{2}}$ and ${D}_{{x}_{3}}$

$| \left(5 , \textcolor{red}{4} , 2\right) , \left(3 , \textcolor{red}{2} , 2\right) , \left(0 , \textcolor{red}{5} , 1\right) | \text{ }$

${D}_{{x}_{2}} = 5 \cdot | \left(2 , 2\right) , \left(5 , 1\right) | - 4 \cdot | \left(3 , 2\right) , \left(0 , 1\right) | + 2 \cdot | \left(3 , 2\right) , \left(0 , 5\right) |$

${D}_{{x}_{2}} = 5 \cdot \left(2 - 10\right) - 4 \cdot \left(3 - 0\right) + 2 \cdot \left(15 - 0\right)$

${D}_{{x}_{2}} = - 40 - 12 + 30 = - 22$

and

$| \left(5 , 3 , \textcolor{red}{4}\right) , \left(3 , 3 , \textcolor{red}{2}\right) , \left(0 , 1 , \textcolor{red}{5}\right) | \text{ }$

${D}_{{x}_{3}} = 5 \cdot | \left(3 , 2\right) , \left(1 , 5\right) | - 3 \cdot | \left(3 , 2\right) , \left(0 , 5\right) | + 4 \cdot | \left(3 , 3\right) , \left(0 , 1\right) |$

${D}_{{x}_{3}} = 5 \cdot \left(15 - 2\right) - 3 \cdot \left(15 - 0\right) + 4 \cdot \left(3 - 0\right)$

${D}_{{x}_{3}} = 65 - 45 + 12 = 32$

Now, according to Cramer's Rule, you have

${x}_{1} = {D}_{{x}_{1}} / D \text{ }$, $\text{ "x_2 = D_(x_2)/D" }$, $\text{ } {x}_{3} = {D}_{{x}_{3}} / D$

This means that you have

${x}_{1} = \frac{2}{2} = \textcolor{g r e e n}{1}$

${x}_{2} = \frac{- 22}{2} = \textcolor{g r e e n}{- 11}$

${x}_{3} = \frac{32}{2} = \textcolor{g r e e n}{16}$