Question

1/2x-1/4y+z=-51/2x-1/3y-1/3z=0x-y+2/3z=-20/3the solution set or no solution inconsistent/dependent or infinite answers inconsistent/dependent

Answer 1

The given system of equations is

`\begin{gathered} (1)/(2)x-(1)/(4)y+z=-5 \\ (1)/(2)x-(1)/(3)y-(1)/(3)z=0 \\ x-y+(2)/(3)z=-(20)/(3) \end{gathered}`

First, let's multiply the second equation by -1 to then add it with the first equation.

`\begin{gathered} (1)/(2)x-(1)/(4)y+z=-5 \\ -(1)/(2)x+(1)/(3)y+(1)/(3)z=0\rightarrow(1)/(2)x-(1)/(2)x-(1)/(4)y+(1)/(3)y+z+(1)/(3)z=-5+0 \\ \end{gathered}`

Now, let's reduce like terms

`\begin{gathered} 0x+(-3y+4y)/(12)+(3z+z)/(3)=-5 \\ (y)/(12)+(4z)/(3)=-5 \end{gathered}`

As you can observe, we've got an equation with two variables, y, and z. Now, we have to find another equation with the same variables. Let's multiply the third equation by -1/2 to then add it with the first equation.

`\begin{gathered} -(1)/(2)x+(1)/(2)y-(2)/(6)z=(20)/(6) \\ (1)/(2)x-(1)/(4)y+z=-5\rightarrow(1)/(2)x-(1)/(2)x+(1)/(2)y-(1)/(4)y+z-(1)/(3)z=(10)/(3)-5 \end{gathered}`

Now, we reduce like terms

`\begin{gathered} 0x+(1)/(4)y+(2)/(3)z=(10-15)/(3) \\ (1)/(4)y+(2)/(3)z=(-5)/(3) \end{gathered}`

Then, we form a new system of two equations with the new equations we found

`\begin{gathered} (y)/(12)+(4z)/(3)=-5 \\ (y)/(4)+(2z)/(3)=-(5)/(3) \end{gathered}`

At this point, let's solve this new system. We have to multiply the second equation by -1/3 to then add them

`\begin{cases}(y)/(12)+(4z)/(3)=-5 \\ -(y)/(12)-(2z)/(9)=(5)/(9)\end{cases}\rightarrow(4z)/(3)-(2z)/(9)=-5+(5)/(9)`

Let's solve for z

`\begin{gathered} (36z-6z)/(27)=(-45+5)/(9) \\ (30z)/(27)=(-40)/(9) \\ z=(-50\cdot27)/(9\cdot30) \\ z=(-1080)/(270) \\ z=-4 \end{gathered}`

Then, we find the value of y using one of the equations in the last system.

`\begin{gathered} (y)/(4)+(2z)/(3)=-(5)/(3) \\ (y)/(4)+(2\cdot(-4))/(3)=-(5)/(3) \\ (y)/(4)-(8)/(3)=-(5)/(3) \\ (y)/(4)=-(5)/(3)+(8)/(3) \\ (y)/(4)=(3)/(3) \\ y=1\cdot4 \\ y=4 \end{gathered}`

At last, we use one of the initial equations to find the value of x.

`\begin{gathered} (1)/(2)x-(1)/(4)y+z=-5 \\ (1)/(2)x-(1)/(4)\cdot4+(-4)=-5 \\ (1)/(2)x-1-4=-5 \\ (1)/(2)x-5=-5 \\ (1)/(2)x=-5+5 \\ x=2\cdot0 \\ x=0 \end{gathered}`

Therefore, the system has the solution set (0, 4, -4), that is, x = 0, y = 4, and z = -4.