# Is it possible to solve the following system of linear equations?

The system is the following: $$-2x + 4y = 7$$ $$3x - 5y = 4$$ i don't usually ask for these kind of things,seems really easy but this really got me. i've tried by many methods; but everytime i check it, it doesn't satisfy one equation. Once i got $Y=15.5$ and $X = -27.5$. As you can see it satisfies the first equation, but not the second one. Is it even possible to solve this?

$(1)$ $$-2x + 4y = 7$$ $(2)$ $$3x - 5y = 4$$

From $(1)$, we get $$2x= 4y-7 \implies x=\frac{4y-7}{2}$$

Substiting into $(2)$:

$$3\cdot\frac{4y-7}{2} -5y = 4$$

$$\implies 3(4y-7)-10y=8$$

$$\implies 12y-10y = 8+21 \implies2y = 29 \implies y=\frac{29}{2}$$

Then $$x = \frac{4\cdot \frac{29}{2}-7}{2} = \frac{2\cdot 29 -7}{2} = \frac{51}{2}$$

These values of $x$ and $y$ satisfy both equations as you would expect.

• Oh, i see, i did it by other method but i missed the sum.; $2y = 21 + 8$ and i did $2y=31$. Commented Mar 23, 2017 at 23:31

Your system is described by the augmented matrix $$\left[\begin{array}{rr|r} -2 & 4 & 7 \\ 3 & -5 & 4 \end{array}\right]$$ Using elementary row operations, we may find the reduced row-echelon form of the system.

First, scale $\DeclareMathOperator{Row}{Row}\Row_1$ by $-1/2$ to obtain $$\left[\begin{array}{rr|r} 1 & -2 & -\frac{7}{2} \\ 3 & -5 & 4 \end{array}\right]$$ Next, add $-3\Row_1$ to $\Row_2$ to get $$\left[\begin{array}{rr|r} 1 & -2 & -\frac{7}{2} \\ 0 & 1 & \frac{29}{2} \end{array}\right]$$ Finally, add $2\Row_2$ to $\Row_1$ to obtain $$\left[\begin{array}{rr|r} 1 & 0 & \frac{51}{2} \\ 0 & 1 & \frac{29}{2} \end{array}\right]$$ This tells us that $x=51/2$ and $y=29/2$.