If X 12y 210 And X 6y 90 Then X

If x + 12y = 210 and x + 6y = 90 then x
Finding the value of x in a pair of linear equations

Introduction

When you encounter two equations that share the same variables, the goal is often to determine the numeric value of one (or both) of those variables. The statement “if x + 12y = 210 and x + 6y = 90 then x” presents exactly that situation: two linear equations in the unknowns x and y. Solving such a system is a fundamental skill in algebra, and it appears in everything from basic homework problems to real‑world modeling in physics, economics, and engineering.

In this article we will walk through the logic behind solving the given pair, demonstrate two classic methods—substitution and elimination—and discuss why the solution makes sense. We’ll also explore variations, common pitfalls, and practical applications so that you can confidently tackle similar problems on your own.

Understanding the Problem

First, let’s rewrite the statements with the missing operators made explicit. The most natural interpretation is:

[ \begin{cases} x + 12y = 210 \ x + 6y = 90 \end{cases} ]

Both equations are linear because each term is either a constant or a variable raised to the first power. The graphs of these equations are straight lines in the xy‑plane, and the solution to the system corresponds to the point where the two lines intersect.

If the lines intersect at a single point, that point gives a unique pair ((x, y)) that satisfies both equations simultaneously. Our task is to isolate x from that intersection.

Method 1: Elimination (Adding/Subtracting Equations)

The elimination method works by adding or subtracting the equations so that one variable cancels out. Because both equations contain x with the same coefficient (1), subtracting the second equation from the first eliminates x immediately:

[ \begin{aligned} (x + 12y) - (x + 6y) &= 210 - 90 \ x + 12y - x - 6y &= 120 \ 6y &= 120 \end{aligned} ]

Now solve for y:

[ y = \frac{120}{6} = 20 ]

With y known, substitute it back into either original equation to find x. Using the simpler second equation:

[ \begin{aligned} x + 6y &= 90 \ x + 6(20) &= 90 \ x + 120 &= 90 \ x &= 90 - 120 \ x &= -30 \end{aligned} ]

Thus, (x = -30) and (y = 20).

Method 2: Substitution

Substitution isolates one variable in one equation and plugs that expression into the other equation. Let’s solve the second equation for x:

[ x = 90 - 6y ]

Now replace x in the first equation:

[ \begin{aligned} (90 - 6y) + 12y &= 210 \ 90 + (-6y + 12y) &= 210 \ 90 + 6y &= 210 \ 6y &= 210 - 90 \ 6y &= 120 \ y &= 20 \end{aligned} ]

Insert y = 20 back into the expression for x:

[ x = 90 - 6(20) = 90 - 120 = -30 ]

Again we obtain (x = -30).

Both methods lead to the same result, confirming the correctness of the solution.

Why the Solution Makes Sense – A Geometric View

Plotting the two lines helps visualize why the intersection occurs at ((-30, 20)).

• Line 1: (x + 12y = 210) → rewrite as (y = \frac{210 - x}{12}).
• Line 2: (x + 6y = 90) → rewrite as (y = \frac{90 - x}{6}).

Both lines have negative slopes (‑1/12 and ‑1/6, respectively), so they descend as x increases. Because the slopes differ, the lines are not parallel and must intersect exactly once. Solving algebraically gave us the intersection point ((-30, 20)). If you substitute x = ‑30 into either line’s equation, you’ll see that the corresponding y value is indeed 20, confirming the geometric interpretation.

Checking the Solution

A good habit is to verify the answer by plugging the found values back into both original equations:

1. (x + 12y = -30 + 12(20) = -30 + 240 = 210) ✔️
2. (x + 6y = -30 + 6(20) = -30 + 120 = 90) ✔️

Both equations hold true, so the solution is correct.

Common Mistakes and How to Avoid Them