• MHB
• Prove It
In summary, the solution to the given system of equations is (5, 8, -6). This can be found by first multiplying the equations by the LCM of the x coefficients, applying Gaussian elimination, and solving for x, y, and z.
Prove It
Gold Member
MHB
Solve the following system for \displaystyle \begin{align*} x, y, z \end{align*}:

\displaystyle \begin{align*} 5\,x - 2\,y + z &= 3 \\ 3\,x + y + 3\,z &= 5 \\ 6\,x + y - 4\,z &= 62 \end{align*}

The LCM of the \displaystyle \begin{align*} x \end{align*} coefficients is 30, so multiplying the first equation by 6, the second by 10 and the third by 5 gives

\displaystyle \begin{align*} 30\,x - 12\,y + 6\,z &= 18 \\ 30\,x + 10\,y + 30\,z &= 50 \\ 30\,x + 5\,y - 20\,z &= 310 \end{align*}

Applying R2 - R1 to R2 and R3 - R1 to R3 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 22\,y + 24\,z &= 32 \\ 17\,y - 26\,z &= 292 \end{align*}

Dividing the second equation by 2 gives

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 11\,y + 12\,z &= 16 \\ 17\,y - 26 \,z &= 292 \end{align*}

The LCM of the y coefficients in rows 2 and 3 is 187, so multiplying the second equation by 17 and the third equation by 11 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ 187\,y - 286\,z &= 3\,212 \end{align*}

Applying R3 - R2 to R2 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ - 490\,z &= 2\,940 \end{align*}

Since \displaystyle \begin{align*} -490\,z = 2\,940 \implies z = -6 \end{align*}, then

\displaystyle \begin{align*} 187\,y + 204 \, \left( -6 \right) &= 272 \\ 187\,y - 1\,224 &= 272 \\ 187\,y &= 1\,496 \\ y &= 8 \end{align*}

and

\displaystyle \begin{align*} 5\,x - 2\,\left( 8 \right) + \left( -6 \right) &= 3 \\ 5\,x - 22 &= 3 \\ 5\,x &= 25 \\ x &= 5 \end{align*}

So the solution is \displaystyle \begin{align*} \left( x , y , z \right) = \left( 5, 8, -6 \right) \end{align*}.

Prove It said:
The LCM of the \displaystyle \begin{align*} x \end{align*} coefficients is 30, so multiplying the first equation by 6, the second by 10 and the third by 5 gives

\displaystyle \begin{align*} 30\,x - 12\,y + 6\,z &= 18 \\ 30\,x + 10\,y + 30\,z &= 50 \\ 30\,x + 5\,y - 20\,z &= 310 \end{align*}

Applying R2 - R1 to R2 and R3 - R1 to R3 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 22\,y + 24\,z &= 32 \\ 17\,y - 26\,z &= 292 \end{align*}

Dividing the second equation by 2 gives

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 11\,y + 12\,z &= 16 \\ 17\,y - 26 \,z &= 292 \end{align*}

The LCM of the y coefficients in rows 2 and 3 is 187, so multiplying the second equation by 17 and the third equation by 11 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ 187\,y - 286\,z &= 3\,212 \end{align*}

Applying R3 - R2 to R2 we have

\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ - 490\,z &= 2\,940 \end{align*}

Since \displaystyle \begin{align*} -490\,z = 2\,940 \implies z = -6 \end{align*}, then

\displaystyle \begin{align*} 187\,y + 204 \, \left( -6 \right) &= 272 \\ 187\,y - 1\,224 &= 272 \\ 187\,y &= 1\,496 \\ y &= 8 \end{align*}

and

\displaystyle \begin{align*} 5\,x - 2\,\left( 8 \right) + \left( -6 \right) &= 3 \\ 5\,x - 22 &= 3 \\ 5\,x &= 25 \\ x &= 5 \end{align*}

So the solution is \displaystyle \begin{align*} \left( x , y , z \right) = \left( 5, 8, -6 \right) \end{align*}.
Correct. The next step would be to write it in form of matrices. 'Gaussian elimination' would be a suitable search key.

Astronuc and Greg Bernhardt

## 1. How do I solve a system of equations?

To solve a system of equations, you first need to have two or more equations with two or more variables. Then, you can use techniques such as substitution, elimination, or graphing to find the values of the variables that satisfy all the equations.

## 2. Can you explain substitution and elimination in solving a system of equations?

In substitution, you solve one equation for one of the variables and substitute that expression into the other equation. This allows you to solve for the remaining variable. In elimination, you manipulate the equations so that when you add or subtract them, one of the variables is eliminated and you can solve for the other variable.

## 3. What is the difference between a consistent and an inconsistent system of equations?

A consistent system of equations has at least one solution that satisfies all the equations. An inconsistent system of equations has no solution that satisfies all the equations. In other words, the lines or curves represented by the equations either intersect at one point or do not intersect at all.

## 4. How do I know if a system of equations has infinite solutions?

If the equations in a system are equivalent, meaning they represent the same line or curve, then the system has infinite solutions. You can also check by graphing the equations and seeing if they overlap completely.

## 5. Can you provide an example of solving a system of equations?

For example, let's say we have the system of equations:
2x + y = 8
3x - y = 2
We can solve this using elimination by multiplying the second equation by 2:
2x + y = 8
6x - 2y = 4
8x = 12
x = 3
Substituting this value back into the first equation, we get:
2(3) + y = 8
6 + y = 8
y = 2
So, the solution to this system of equations is x = 3 and y = 2.

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