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Algebra · real student question

Solve the system of equations y = 4x - 2 and y = x + 3.

Question

Solve the system

{y=4x−2y=x+3\begin{cases} y=4x-2\\ y=x+3 \end{cases}

Step-by-step solution

  1. Notice both equations are already solved for yy. That makes substitution trivial: whatever xx is, the two right-hand sides must produce the same yy, so they equal each other. No elimination or matrix work is needed.

  2. Set the right-hand sides equal and collect the xx terms.

    4x−2=x+3 ⟹ 4x−x=3+2 ⟹ 3x=5 ⟹ x=534x-2=x+3\ \Longrightarrow\ 4x-x=3+2\ \Longrightarrow\ 3x=5\ \Longrightarrow\ x=\frac{5}{3}

  3. Substitute back to get yy. Using the simpler line y=x+3y=x+3:

    y=53+3=53+93=143y=\frac{5}{3}+3=\frac{5}{3}+\frac{9}{3}=\frac{14}{3}

  4. Confirm with the other equation. The point must satisfy y=4x−2y=4x-2 as well:

    4(53)−2=203−63=143 ✓4\left(\frac{5}{3}\right)-2=\frac{20}{3}-\frac{6}{3}=\frac{14}{3}\ \checkmark

    Both lines give the same yy, so (53,143)\left(\tfrac53,\tfrac{14}3\right) is the intersection point.

  5. Interpret the answer graphically. The gradients 44 and 11 are different, so the lines are not parallel and there is exactly one intersection — which is why a single point, and not a line or the empty set, is the correct form of the answer. In decimals it sits at about (1.67, 4.67)(1.67,\,4.67), which is what a graphing tool would show.

Answer

x=53,y=143x=\frac{5}{3},\qquad y=\frac{14}{3}

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