Solve System of Equations 2y x - 4 and 4x 3y 5 With Substitution and Elimination

The system written as 2y = x − 4 and 4x + 3y = 5 solves to x = 2 and y = −1, or the point (2, −1). Substituting x = 2y + 4 into the second equation gives 8y + 16 + 3y = 5, so 11y = −11 and y = −1. Checking both equations confirms the pair satisfies each one.

What the notation actually means

The keyword phrase drops operators that algebra normally prints. Read "2y x - 4" as 2y = x − 4, a linear equation relating y to x. Read "4x 3y 5" as 4x + 3y = 5, a second linear equation in the same two variables. Solving the system means finding every ordered pair (x, y) that satisfies both statements at once.

Each equation is a straight line in the coordinate plane. A solution is a point lying on both lines. Two distinct nonparallel lines cross exactly once, so this system has exactly one solution. That geometric fact drives everything that follows.

Rearranging the first equation into slope-intercept form gives y = (x − 4)/2, a line with slope 0.5 and y-intercept −2. The second becomes y = (5 − 4x)/3, a line with slope −4/3 and y-intercept 5/3. Setting the right sides equal reproduces the same algebra: (x − 4)/2 = (5 − 4x)/3.

This is why the problem type matters. Linear systems have predictable solution counts determined by slopes, not by how large the coefficients look. A single crossing point means the answer is unique, and any correct method must return the same pair.

Mechanism: why substitution works

Substitution exploits a logical rule: if two quantities are equal, either can replace the other in any expression without changing its value. The first equation certifies that x − 4 and 2y are the same number. So anywhere x appears, you may swap in an equivalent expression.

Solve the first equation for x to get x = 2y + 4. This isolates one variable in terms of the other. Substitute that expression into the second equation: 4(2y + 4) + 3y = 5. The equation now contains only y, so it becomes solvable in one unknown.

Expand to 8y + 16 + 3y = 5, combine like terms to 11y + 16 = 5, subtract 16 to get 11y = −11, then divide by 11 to reach y = −1. Back-substitute into x = 2y + 4 to get x = 2(−1) + 4 = 2. The pair (2, −1) is the candidate solution.

Verification is not optional decoration. Plug into the original first equation: 2(−1) = −2 and 2 − 4 = −2, so both sides match. Plug into the second: 4(2) + 3(−1) = 8 − 3 = 5, matching the right side. Both checks pass, so the solution is confirmed.

The deeper reason substitution is safe: it preserves the solution set. Replacing a variable with an equal expression never adds or removes solutions, provided the expression is defined everywhere on the relevant domain. For linear equations it always is, so no extraneous solutions can appear.

Quantified context: elimination as a cross-check

Elimination attacks the same problem from a different angle, and comparing outputs is a professional habit. Align the equations: x − 2y = 4 and 4x + 3y = 5. Multiply the first equation by 4 to get 4x − 8y = 16. Subtract the second equation from this result: (4x − 8y) − (4x + 3y) = 16 − 5, which simplifies to −11y = 11, so y = −1.

Substitute y = −1 into x − 2y = 4 to get x + 2 = 4, so x = 2. Both methods agree, which is strong evidence the arithmetic is correct. When two independent procedures converge on the same pair, the probability of a matching error is small.

The determinant of the coefficient matrix quantifies whether a unique solution exists. For the system x − 2y = 4 and 4x + 3y = 5, the determinant is (1)(3) − (−2)(4) = 3 + 8 = 11. A nonzero determinant of 11 guarantees exactly one solution, and it explains the 11 that appeared in both methods.

That is not a coincidence. Cramer's rule expresses each variable as a ratio of determinants with the coefficient determinant in the denominator. Here x = 22/11 = 2 and y = −11/11 = −1. The shared 11 is the system's structural fingerprint, and it appears regardless of which technique you choose.

A quick sanity check on magnitude helps too. Slopes of 0.5 and −4/3 are far apart, so the lines cross at a shallow angle near the origin. A crossing point at (2, −1) sits close to where both intercepts suggest, which is consistent with the computed answer.

Edge cases that change the answer count

The unique-solution outcome depends on the lines being distinct and nonparallel. Three other cases exist, and each changes the count of solutions rather than the arithmetic.

Parallel distinct lines share a slope but not an intercept. They never meet, so the system has no solution and is called inconsistent. Substitution or elimination produces a false statement such as 0 = 7, which signals the contradiction.

Coincident lines are the same line written twice. Every point on the line satisfies both equations, so there are infinitely many solutions. Elimination produces a true but uninformative statement such as 0 = 0, which signals dependence.

Nonlinear systems introduce their own edge cases. Substituting a line into a circle can yield zero, one, or two intersection points, and substitution can generate extraneous roots when squaring occurs. Always verify candidate solutions against the original equations, not the manipulated ones.

A professional caveat: elimination by subtraction can silently drop a factor if you divide both sides by an expression that might be zero. In linear systems the coefficient is a constant, so this risk is minimal, but in systems with variable coefficients it becomes real.

Misconceptions and red-flag warnings

Several errors recur often enough to deserve explicit warning. Each produces a wrong answer that looks plausible, which is what makes them dangerous.

Substituting into the same equation you solved instead of the other one, which yields a tautology and hides the real constraint. Dropping a negative sign when rearranging x − 2y = 4, producing x = 2y − 4 instead of x = 2y + 4. Adding equations when the goal was to eliminate a variable, when subtraction was required because the coefficients already matched. Skipping verification, which lets arithmetic slips survive to the final answer. Treating "no solution" and "infinitely many solutions" as the same failure, when they describe opposite geometries.

One counterintuitive detail: the system's answer does not depend on which variable you isolate first. Isolating y instead of x gives y = (x − 4)/2, and substituting into 4x + 3y = 5 produces 4x + (3x − 12)/2 = 5, which clears to 11x = 22 and x = 2. Different path, identical destination.

Another misconception is that larger coefficients mean a harder or different problem. The method and the answer count depend on the determinant, not on the size of the numbers. A system with coefficients in the thousands can still have a single tidy solution.

Comparison of solution methods

Each method suits different situations, and choosing well saves time and reduces error. The table compares four standard approaches on the two-equation, two-variable case.

Substitution and elimination both deliver exact answers and should agree. Graphing is best treated as a sanity check, because hand-read intersections rarely give exact values. Matrix methods pay off when you solve many systems with the same structure.

For the system in this article, elimination is the faster route because multiplying one equation by 4 aligns the x coefficients cleanly. Substitution is equally valid and arguably clearer for a first exposure, since it isolates one variable explicitly.

Practical actions and thresholds

Follow a fixed checklist so the procedure becomes mechanical and verifiable. The steps below apply to any two-by-two linear system, including this one.

Step one: rewrite both equations in standard form, ax + by = c, so coefficients align. Here that gives x − 2y = 4 and 4x + 3y = 5. Standard form makes the determinant easy to read.

Step two: compute the determinant ad − bc. If it is nonzero, expect exactly one solution and proceed. If it is zero, check whether the lines coincide or are parallel before doing more work.

Step three: choose substitution when a coefficient is 1 or −1, and elimination when coefficients share a small common multiple. For this system, the coefficient of x is 1, so substitution is natural.

Step four: solve, then back-substitute to find the second variable. Step five: verify in both original equations. A single failed check means an arithmetic error somewhere earlier.

Step six: state the answer as an ordered pair and, if relevant, as a geometric point. Reconcile the two methods if you have time; agreement raises confidence substantially.

Threshold rule: treat a determinant of exactly zero as a stop sign, not a minor inconvenience. Recompute the determinant from the standard-form coefficients before concluding anything about the solution count.

Limits of what this establishes

This worked example demonstrates exact solution of one small linear system. It does not generalize automatically to larger systems, where Gaussian elimination with pivoting becomes the standard tool for numerical stability.

Floating-point arithmetic introduces rounding error that exact fractions avoid. In applied settings, a determinant near zero can amplify small input errors into large output errors, a condition called ill-conditioning. The determinant of 11 here is comfortably away from zero.

Real-world models often have more equations than unknowns or noisy measurements, so no exact solution exists and least-squares fitting replaces direct solving. That is a different problem class with different guarantees.

The methods shown assume real coefficients and real solutions. Complex solutions, constrained domains, and integer-only requirements each add separate conditions that this example does not address. State those assumptions explicitly when presenting results.

Finally, the answer (2, −1) is correct for the system as interpreted. If the original notation meant something other than 2y = x − 4 and 4x + 3y = 5, the solution changes entirely. Confirm the intended operators before trusting any computed pair.