Chapter 3.5, Problem 4CYU

To determine

To calculate: The solution set of system of equations,

  $\begin{array}{c}-4r-s+3t=-9\\ 3r+2s-t=3\\ r+3s-5t=29\end{array}$

Answer to Problem 4CYU

The solution set of the system of equations is $\left(-2,2,-5\right)$.

Explanation of Solution

Given information:

The system of equations is provided below,

  $\begin{array}{c}-4r-s+3t=-9\\ 3r+2s-t=3\\ r+3s-5t=29\end{array}$

Formula used:

A system of equations with three variables can be solved by elimination method and substitution method. Select any two pairs of equations from the system and eliminate the same variable by adding or subtracting and then simplify it.

Calculation:

Consider the provided system of equations as,

  $\begin{array}{c}-4r-s+3t=-9\\ 3r+2s-t=3\\ r+3s-5t=29\end{array}$

Recall that a system of equations with three variables can be solved by elimination method and substitution method. Select any two pairs of equations from the system and eliminate the same variable by adding or subtracting and then simplify it.

So, multiply the equation $-4r-s+3t=-9$ by 2 and then add second and the first equation to eliminate s as,

  $\begin{array}{c}2\left(-4r-s+3t\right)=2\left(-9\right)\\ -8r-2s+6t=-18\end{array}$

Now add the second and the first equation as,

  $\begin{array}{c}3r+2s-t=3\\ (+)-8r-2s+6t=-18\\ -5r+5t=-15\end{array}$

Multiply the equation $-4r-s+3t=-9$ by 3 and then add first and the third equation to eliminate s as,

  $\begin{array}{c}3\left(-4r-s+3t\right)=3\left(-9\right)\\ -12r-3s+9t=-27\end{array}$

Now add the third and the first equation as,

  $\begin{array}{c}-12r-3s+9t=-27\\ (+)r+3s-5t=29\\ -11r+4t=2\end{array}$

Multiply $-5r+5t=-15$ by $-4$ and $-11r+4t=2$ by 5 and then add both the equations.

  $\begin{array}{c}-4\left(-5r+5t\right)=-4\left(-15\right)\\ 20r-20t=60\end{array}$

  $\begin{array}{c}5\left(-11r+4t\right)=5\left(2\right)\\ -55r+20t=10\end{array}$

Add both the equations to eliminate t and get the value of r ,

  $\begin{array}{c}20r-20t=60\\ (+)-55r+20t=10\\ -35r=70\end{array}$

Simplify it further as,

  $\begin{array}{c}-35r=70\\ r=\frac{70}{-35}\\ r=-2\end{array}$

Substitute $r=-2$ in $-5r+5t=-15$ and solve it to get the value of t as,

  $\begin{array}{c}-5\left(-2\right)+5t=-15\\ 10+5t=-15\\ 5t=-25\\ t=-5\end{array}$

Now, substitute $r=-2$ and $t=-5$ in equation third to get the value of s as,

  $\begin{array}{c}-2+3s-5\left(-5\right)=29\\ 3s-2+25=29\\ 3s=6\\ s=2\end{array}$

To check whether the solution set is correct, substitute the values of r , s and t in second equation as,

  $\begin{array}{c}3r+2s-t=3\\ 3\left(-2\right)+2\left(2\right)-\left(-5\right)=3\\ -6+4+5=3\\ 3=3\end{array}$

Since, left-hand side is equal to right-hand side, the solution set is true for all the equations.

Thus, the solution set of the system of equations is $\left(-2,2,-5\right)$.