Impossible simultaneous equation?

**Potrik98** · September 18th 2012, 06:35 AM

Hey! I just got this equation in my math homework. I couldnt solve it, and noone else could.
Is this an impossible equation? and if so, why?
Please answer fast.

Equation:

I: -6x + 3y = 5

II: -16x + 3y = 7

**MarkFL** · September 18th 2012, 06:41 AM

There is a solution.

**Potrik98** · September 18th 2012, 06:52 AM

Originally Posted by MarkFL2

There is a solution.

Whats the solution? I didnt find any.

**MarkFL** · September 18th 2012, 06:55 AM

Try solving both equations for 3y, then equate the results to find x, then using either equation to determine y.

**Soroban** · September 18th 2012, 08:41 AM

Hello, Potrik98!

Hey! I just got this problem in my math homework.
I couldn't solve it, and no one else could.
. . Really? No one?

Is this an impossible equation? .And if so, why?
Please answer fast.

Equations: . $\begin{array}{cccc}-6x + 3y &=& 5 & [1] \\-16x + 3y &=& 7 & [2] \end{array}$

There are several method for solving system of linear equations.
Which ones are you familiar with?
It seems that you and your classmates have learned NONE of them.

Elimination

We have: . $\begin{array}{cccc}\text{-}6x + 3y &=& 5 & [1] \\\text{-}16x + 3y &=& 7 & [2] \end{array}$

Subtract [1]-[2]: . $10x \:=\:\text{-}2 \quad\Rightarrow\quad \boxed{x \:=\:\text{-}\tfrac{1}{5}}$

Substitute into [1]: . $\text{-}6\left(\text{-}\tfrac{1}{5}\right) + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:\tfrac{19}{5} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}$

Substitution

Solve [1] for $y\!:\;\text{-}6x + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:6x + 5 \quad\Rightarrow\quad y \:=\:2x+\tfrac{5}{3}$

Substitute into [2]: . $\text{-}16x + 3(2x + \tfrac{5}{3}) \:=\:7 \quad\Rightarrow\quad \text{-}16x + 6x + 5 \:=\:7$

. . . . . . . . . . . . . . . $\text{-}10x \:=\:2 \quad\Rightarrow\quad \boxed{x \:=\:\text{-}\tfrac{1}{5}}$

Substitute into [1]: . $\text{-}6\left(\text{-}\tfrac{1}{5}\right) + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:\tfrac{19}{5} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}$

Cramer's Rule

We have: . $\left|\begin{array}{cc|c} \text{-}6 & 3 & 5 \\ \text{-}16 & 3 & 7 \end{array}\right|$

$D \;=\;\begin{vmatrix}-6&3\\-16&3\end{vmatrix} \:=\:(\text{-}18)-(\text{-}48) \:=\:30$

$D_x \:=\:\begin{vmatrix} 5&3\\7&3\end{vmatrix} \:=\:15-21 \:=\:\text{-}6$

. . $x \;=\;\frac{D_x}{D} \;=\;\frac{\text{-}6}{30} \quad\Rightarrow\quad \boxed{x\;=\;\text{-}\tfrac{1}{5}}$

$D_y \:=\:\begin{vmatrix}\text{-}6&5\\\text{-}16&7\end{vmatrix} \:=\:\text{-}42+80 \:=\:38$

. . $y \;=\;\frac{D_y}{D} \;=\;\frac{38}{30} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}$

**Potrik98** · September 18th 2012, 09:28 PM

Thanks for the help, though I misstyped the equation, what I meant was:

I: -6x + 3y = 5

II: -16x + 8y = 7

the + 3y in equation II was really a +8y, though i will try some of the methods you listed

**Prove It** · September 18th 2012, 09:37 PM

Originally Posted by Potrik98

Thanks for the help, though I misstyped the equation, what I meant was:

I: -6x + 3y = 5

II: -16x + 8y = 7

the + 3y in equation II was really a +8y, though i will try some of the methods you listed

To use elimination you need one of the variables to have the same coefficients. So multiply each equation by a number which will give the LCM of 6 and 16 (if you want to eliminate x), or multiply each equation by a number which will give the LCM of 3 and 8 (if you want to eliminate y).

**Potrik98** · September 18th 2012, 10:32 PM

Well im still having problems:

[I] -6x + 3y = 5
[II] -16x + 8y = 7

Elimination:

[I] -6x + 3y = 5 | * 8
[II] -16x + 8y = 7 | * -3

[I] -48x + 24y = 40
[II] 48x - 24y = -21

0x + 0y = 19

???
How is this possible?
???

Substutition:

[I] -6x + 3y = 5
[II] -16x + 8y = 7

[I]
-6x + 3y = 5
3y = 5 + 6x | : 3
y = 5/3 + 2x

[II]
-16x + 8y = 7
-16x + 8(5/3 + 2x) = 7
-16x + 40/3 + 16x = 7
-16x + 16x = 7 - 40/3
0x = 7 - 40/3
0x = -19/3

???

This is completely strange. I've tried to slove the equation on several simoultaneous equation solving sites, and my Casio fx-9860 GII calculator, and they all give me the same answer.

Mathemathical error
Solution not found

Please help!

**MarkFL** · September 18th 2012, 10:45 PM

The two equations represent lines in the Cartesian plane...lines with equal slope but different intercepts, hence they are parallel and distinct, and will never cross, so there is no solution.

**Vlasev** · September 18th 2012, 11:33 PM

Originally Posted by Potrik98

0x + 0y = 19

???
How is this possible?
???

Exactly. It is not possible, so the two equations do not have solutions. If you ever run into the same situation again, you can safely say that the equations have no simultaneous solutions.

**MarkFL** · September 19th 2012, 01:01 AM

Suppose you have two linear equations where none of the coefficients of the variables are zero:

$A_1x+B_1y=C_1$

$A_2x+B_2y=C_2$

A quick and easy check you may perform first to tell you of the nature of the solution is as follows:

If $\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}$ then there are an infinite number of solutions, as they are the same equation.

If $\frac{A_1}{A_2}=\frac{B_1}{B_2}\ne\frac{C_1}{C_2}$ then there is no solution, as they are separate, but parallel lines.

If $\frac{A_1}{A_2}\ne\frac{B_1}{B_2}$ then there is a unique solution.

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