Question

Show that the equations x+y+z = 4, 2x+5y-2z =3, x+7y-7z =5 are not consistent

Answer 1

We can start by using the second equation to eliminate x:

2x + 5y - 2z = 3

2x = -5y + 2z + 3

x = (-5/2)y + z + 3/2

Now we can substitute this expression for x into the first and third equations:

x + y + z = 4

(-5/2)y + z + 3/2 + y + z = 4

(-5/2)y + 2z = 5/2

x + 7y - 7z = 5

(-5/2)y + z + 3/2 + 7y - 7z = 5

(9/2)y - 6z = 7/2

Now we have a system of two equations with two variables, (-5/2)y + 2z = 5/2 and (9/2)y - 6z = 7/2. We can use any method to solve for y and z, such as substitution or elimination. However, we will find that the system is inconsistent, meaning there is no solution that satisfies both equations.

Multiplying the first equation by 9 and the second equation by 5 and adding them, we get:

(-45/2)y + 18z = 45/2

(45/2)y - 30z = 35/2

Adding these two equations, we get:

-12z = 40/2

-12z = 20

z = -5/3

Substituting z = -5/3 into (-5/2)y + 2z = 5/2, we get:

(-5/2)y + 2(-5/3) = 5/2

(-5/2)y - 10/3 = 5/2

(-5/2)y = 25/6

y = -5/12

Substituting y = -5/12 and z = -5/3 into any of the original equations, we get:

x + y + z = 4

x - 5/12 - 5/3 = 4

x = 29/12

Therefore, the solution is (x, y, z) = (29/12, -5/12, -5/3). However, if we substitute these values into any of the original equations, we will find that it does not satisfy the equation. For example:

2x + 5y - 2z = 3

2(29/12) + 5(-5/12) - 2(-5/3) = 3

29/6 - 5/2 + 5/3 ≠ 3

Since there is no solution that satisfies all three equations, the system is inconsistent.

Explanation:

Answer 2

See below for proof.

Explanation:

A system of equations is not consistent when there is no solution or no set of values that satisfies all the equations simultaneously. In other words, the equations are contradictory or incompatible with each other.

Given system of equations:

`\begin{cases}x+y+z = 4\\2x+5y-2z =3\\x+7y-7z =5\end{cases}`

Rearrange the first equation to isolate x:

`x=4-y-z`

Substitute this into the second equation to eliminate the term in x:

`\begin{aligned}2x+5y-2z&=3\\2(4-y-z)+5y-2z&=3\\8-2y-2z+5y-2z&=3\\-2y-2z+5y-2z&=-5\\5y-2y-2z-2z&=-5\\3y-4z&=-5\end{aligned}`

Subtract the first equation from the third equation to eliminate x:

`\begin{array}{cccrcrcl}&x&+&7y&-&7z&=&5\\\vphantom{\frac12}-&(x&+&y&+&z&=&4)\\\cline{2-8}\vphantom{\frac12}&&&6y&-&8z&=&1\end{aligned}`

Now we have two equations in terms of the variables y and z:

`\begin{cases}3y-4z=-5\\6y-8z=1\end{cases}`

Multiply the first equation by 2 so that the coefficients of the variables of both equations are the same:

`\begin{cases}6y-8z=-10\\6y-8z=1\end{cases}`

Comparing the two equations, we can see that the coefficients of the y and z variables are the same, but the numbers they equate to is different. This means that there is no way to add or subtract the equations to eliminate one of the variables.

For example, if we subtract the second equation from the first equation we get:

`\begin{array}{crcrcl}&6y&-&8z&=&-10\\\vphantom{\frac12}-&(6y&-&8z&=&\:\:\;\;\:1)\\\cline{2-6}\vphantom{\frac12}&&&0&=&-11\end{aligned}`

Zero does not equal negative 11.

Since we cannot eliminate the variable y or z, we cannot find a unique solution that satisfies all three equations simultaneously. Therefore, the system of equations is inconsistent.