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
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: .$\displaystyle \begin{array}{cccc}-6x + 3y &=& 5 & [1] \\-16x + 3y &=& 7 & [2] \end{array}$
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: .$\displaystyle \left|\begin{array}{cc|c} \text{-}6 & 3 & 5 \\ \text{-}16 & 3 & 7 \end{array}\right|$
$\displaystyle D \;=\;\begin{vmatrix}-6&3\\-16&3\end{vmatrix} \:=\:(\text{-}18)-(\text{-}48) \:=\:30$
$\displaystyle D_x \:=\:\begin{vmatrix} 5&3\\7&3\end{vmatrix} \:=\:15-21 \:=\:\text{-}6$
. . $\displaystyle x \;=\;\frac{D_x}{D} \;=\;\frac{\text{-}6}{30} \quad\Rightarrow\quad \boxed{x\;=\;\text{-}\tfrac{1}{5}}$
$\displaystyle D_y \:=\:\begin{vmatrix}\text{-}6&5\\\text{-}16&7\end{vmatrix} \:=\:\text{-}42+80 \:=\:38$
. . $\displaystyle y \;=\;\frac{D_y}{D} \;=\;\frac{38}{30} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}$
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).
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!
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.