There are a few different ways to solve a system of equations. One is substitution in which you need to transform one of the equations into y =, or x = form. Once done you can substitute in that value of y or x into the second equation and solve for one variable, and then with that answer you'll be able to solve for the other variable. For the first system of equations:
{ y=6x-2
{12x-2y=5
The top equation is already in y = form. So the next step is substituting in '6x - 2' for 'y' in the second equation.
That gives a second equation of 12x - 2(6x - 2) = 5
Distribute (negative) -2 into the '6x - 2'. We distribute a -2 instead of (positive) 2 to take care of the minus sign at the same time we distribute. This leaves the second equation as: 12x -12x + 4 = 5
Combining like terms gives 0 + 4 = 5, or 4 = 5
This of course is not true, which means this system of equations has no solution. If you were to graph each of these lines, you would see that they never cross (both lines crossing at a point would indicate the system of equations has a solution). This system of equations does not have a solution.
For the second set of equations, we can solve by elimination! This is where you create a common coefficient between both equations, have at least one of the variables cancel out, and then solve for the remaining variable. Once you have this solution, you can solve for the variable you initially cancelled out. Here are the equations:
{2x+3y=-6
{5x-2y=23
Since the coefficients of y have opposite signs (one positive, one negative), we will multiply each equation by a number so that the y coefficients end up being the same number. The least common multiple of 3 and 2 is 6. So we multiply the entire top equation by 2, and the entire bottom equation by 3. This gives a new set of equations:
4x + 6y = -12
15x - 6y = 69
We can add these equations together, which will make the 'y' variables cancel out, leaving:
19x = 57
Dividing both sides by 19 gives x = 3.
And since we know x = 3, we can substitute that 3 in for 'x' in any equation we've used thus far (the original system of equations and the ones we created).
It seems simplest to input the 3 into the very first equation of the system since its number values are smallest: 2x + 3y = -6
Substituting 3 in for x gives: 2(3) + 3y = -6, or 6 + 3y = -6
Subtract 6 from both sides to give: 3y = -12
Lastly divide both sides by 3, to give y = -4
So for this second set of equations, there is a solution, which is x = 3, y = -4
