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

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- Sep 18th 2012, 06:35 AMPotrik98Impossible simultaneous equation?
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 - Sep 18th 2012, 06:41 AMMarkFLRe: Impossible simultaneous equation?
There is a solution.

- Sep 18th 2012, 06:52 AMPotrik98Re: Impossible simultaneous equation?
- Sep 18th 2012, 06:55 AMMarkFLRe: Impossible simultaneous equation?
Try solving both equations for 3

*y*, then equate the results to find*x*, then using either equation to determine*y*. - Sep 18th 2012, 08:41 AMSorobanRe: Impossible simultaneous equation?
Hello, Potrik98!

Quote:

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

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

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

Substitute into [1]: .$\displaystyle \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 $\displaystyle y\!:\;\text{-}6x + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:6x + 5 \quad\Rightarrow\quad y \:=\:2x+\tfrac{5}{3}$

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

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

Substitute into [1]: .$\displaystyle \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: .$\displaystyle \left|\begin{array}{cc|c} \text{-}6 & 3 & 5 \\ \text{-}16 & 3 & 7 \end{array}\right|$

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

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

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

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

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

- Sep 18th 2012, 09:28 PMPotrik98Re: Impossible simultaneous equation?
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 - Sep 18th 2012, 09:37 PMProve ItRe: Impossible simultaneous equation?
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).

- Sep 18th 2012, 10:32 PMPotrik98Re: Impossible simultaneous equation?
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! - Sep 18th 2012, 10:45 PMMarkFLRe: Impossible simultaneous equation?
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.

- Sep 18th 2012, 11:33 PMVlasevRe: Impossible simultaneous equation?
- Sep 19th 2012, 01:01 AMMarkFLRe: Impossible simultaneous equation?
Suppose you have two linear equations where none of the coefficients of the variables are zero:

$\displaystyle A_1x+B_1y=C_1$

$\displaystyle A_2x+B_2y=C_2$

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

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

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

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