Method 1 (Substitution): One way to solve this is to take the first equation of the system, x−4y=6, and solve it for x in terms of y to get x=4y+6. Then plug this into the second equation to make 3x+4y=10 become 3(4y+6)+4y=10, which simplifies to 12y+18+4y=10, then 16y=−8, then y=−12.
Now use the equation x=4y+6 to find x: x=4⋅(−12)+6=−2+6=4.
The answer is the point (x,y)=(4,−12).
(BTW, all this work implies the only possible solution is (x,y)=(4,−12) (so if it is a solution, it is unique). You can confirm it truly is a solution by substitution back into the original system (checking your work proves it is a solution).)
Method 2 (Elimination): Notice that the −4y in the first equation and the 4y in the second equation will cancel if we add the two equations. Do so to get: 4x=16 so x=4. Then substitute this into either of the original equations to get, for instance 4−4y=6 so 4y=−2 and y=−12, resulting in (x,y)=(4,−12).
A picture of this situation is shown below. The equation x−4y=6 is equivalent to y=14x−32 and is the red line with a slope of 14 and a y-intercept of −32. The equation 3x+4y=10 is equivalent to y=−34x+52 and is the blue line with a slope of −34 and a y-intercept of 52.
![enter image source here]()
