Question

solve using elimination
4x-7y=15
-8x +14y=-30

Answer 1

no solution... once you multiply the first eq by 2 in order to make the second eq terms match the first one, everything cancels out

Answer 2

(2)(4x-7y=15) => 8x-14y=30

Answer 3

that's what I meant lol

Answer 4

Your system of equations:
\[\begin{array}{ccc} 4x - 7y& = & 15 \\ -8x + 14y & = & -30 \end{array}\]
As @kl0723 mentioned, you can multiply the first equation by 2 and add to the second equation in order to change it. Your row operation would thus be
\[\text{eqn. 2} = \left( 2* \text{eqn1} \right)+ \text{eqn2}\]
Keep in mind that the first equation wasn't changed, only the second equation was changed.
The system of equations then becomes:
\[\begin{array}{rcl} 4x - 7y& = & 15 \\ 0x + 0y & = & 0 \end{array}\]
Looking at the second equation, we see that
\[ 0x + 0y = 0\]
This is true for any pair of (x,y), so we have infinitely many solutions (the system is consistent) as opposed to what was mentioned in another post above, which says the system has no solution (meaning the system is inconsistent).
To find a possible point, solve one of the equations for y. I'll solve the first equation:
\[4x - 7y =15\]
\[y = \frac{4x-15}{7}\]
\[\left( x, \frac{4x-15}{7} \right)\]
^ From this, we can choose an x and get a corresponding y which will satisfy the system of equations you provided.