Question

Stuck! Help Me! Solve the following system using the substitution method.

16x + 3y = 0
8x + y = 1

Answer 1

\[(1)\quad 16x + 3y = 0 \\
(2)\quad 8x + y = 1\]

From 2 you can see that \(\Large y = 1 - 8x\), substitute that in 1 to get:
\[16x + 3(1 - 8x) = 0\\
16x + 3 - 24x = 0\\
-8x = -3\\
x = \frac{3}{8}\]
then substitute x = 3/8 in any of the equations to find y:
\[8\left(\frac{3}{8}\right)+y = 1\\
y = -2\]

Then the solution is \(\Large (x, y ) = \left(\frac{3}{8}, -2\right)\)

Answer 2

Thank you :)