Question

If you were to solve the system using the addition method, what would you multiply the first equation by in order to make the y’s add out of this system?

-2x + 3y = 6
7x – 6y = 8

2

-2

4

none of the above

Answer 1

there is -6y in 2nd equation,
to make the '+3y' in fist equation as '+6y' , what will you multiply it ?

Answer 2

2?

Answer 3

yes, thats correct :)

Answer 4

Thanks!

Answer 5

welcome ^_^

Answer 6

\[\begin{align}-2x + 3y &= 6\qquad(i)\\7x – 6y &= 8\qquad(ii)\\\\-4x + 6y &= 12\qquad(2i)\end{align}\]now adding \((ii)+(2i)\) we get
\[\begin{align}3x&=20\\ x&=\frac{20}3\\&=6+\frac23\end{align}\]
we can substitute this back into \((i)\) to get
\[\begin{align}-2\left(\frac{20}3\right) + 3y &= 6\\3y&=6+\frac{40}3\\y&=2+\frac{40}9\\&=6+\frac49\end{align}\]

and we have arrived at the solution to the system
\[(x,y)=\left(6\tfrac23,6\tfrac49\right)\]