It's a common sight in math classrooms: a pair of equations, each with two variables, and the challenge to find the value of one specific variable, like 'x'. Let's take a look at the system you've presented: x - 12y = -210 and x - 6y = 90. Our goal is to isolate 'x'.
Think of these equations as two different statements about the relationship between 'x' and 'y'. We have two pieces of information, and we need to use them together to figure out what 'x' must be.
One straightforward way to tackle this is through elimination. Notice that both equations have an 'x' term with a coefficient of 1. This makes it easy to eliminate 'x' if we subtract one equation from the other. Let's try subtracting the first equation (x - 12y = -210) from the second equation (x - 6y = 90).
(x - 6y) - (x - 12y) = 90 - (-210)
When we distribute the negative sign in the first part, we get: x - 6y - x + 12y = 90 + 210
See how the 'x' terms cancel each other out? That's the beauty of elimination!
Now we're left with: 6y = 300
From this, we can easily find the value of 'y': y = 300 / 6 y = 50
Great! We've found 'y'. But our original question was about 'x'. Now that we know 'y' is 50, we can substitute this value back into either of the original equations to solve for 'x'. Let's use the second equation, x - 6y = 90, because the numbers seem a bit simpler.
x - 6(50) = 90
Now, perform the multiplication: x - 300 = 90
To get 'x' by itself, we add 300 to both sides of the equation: x = 90 + 300 x = 390
And there we have it! The value of 'x' in this system of equations is 390.
It's fascinating how two simple linear equations can hold a unique solution, and by using methods like elimination or substitution, we can systematically uncover those hidden values. It's like piecing together a puzzle, where each equation gives us a crucial clue.