Solve A System Of Linear Equations Using Substitution Method 2x Y 5 X Y 1

Learn step-by-step how to solve the system of linear equations 2x + y = 5 and x + y = 1 using the substitution method. Includes practical examples and common pitfalls.

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Understanding the Substitution Method

The substitution method solves a system of linear equations by isolating one variable in one equation and substituting it into the other. For the system 2x + y = 5 and x + y = 1, start by solving the second equation for x: x = 1 - y. Substitute this into the first equation to get 2(1 - y) + y = 5, which simplifies to 2 - 2y + y = 5, then -y = 3, so y = -3. Now substitute y back: x = 1 - (-3) = 4. The solution is x = 4, y = -3.

Key Steps in the Substitution Process

First, choose the simpler equation to solve for one variable—here, the second equation for x. Substitute carefully to avoid errors in distribution, like multiplying the 2 in 2(1 - y). Simplify algebraically by combining like terms, solve for the remaining variable, and back-substitute to find the other. Verify by plugging values back into both originals: 2(4) + (-3) = 5 and 4 + (-3) = 1, both true.

Practical Example with Graphing Insight

Consider this system as lines intersecting at (4, -3). If you graph y = 5 - 2x and y = 1 - x, they cross at x=4, y=-3. In practice, like budgeting where 2x + y = 5000 (groceries and rent) and x + y = 3000 (total), substitution quickly finds x=2000, y=1000 without graphing tools.

Applications and Common Misconceptions

This method is vital in engineering for circuit analysis or economics for supply-demand models. A misconception is that substitution only works for simple equations—it's versatile for any solvable system. Avoid assuming no solution without checking; here, it's consistent. It promotes algebraic precision over trial-and-error.

Frequently Asked Questions

What if the system has no solution using substitution?
Can I substitute for y instead of x?
Why verify the solution after substitution?