A correct solution begins with equations that describe the quantities and units in the prompt. Then check the pair and the expression requested.
A system of two linear equations describes two conditions that must hold at the same time. Many errors occur before solving: variables are undefined, a total is attached to the wrong quantity, or a fixed cost is treated as a per-item rate.
Translate before choosing a solving method
Define each variable with its unit. Write one equation for each independent condition, then read the equations back in words. A quantity total uses counts; a cost total uses price multiplied by count. For mixtures, distinguish the total amount from the amount of a component.
Worked example: quantities and costs
This and the other examples below are original teaching exercises, not released SAT questions.
A school buys 18 notebooks: some cost $3 each and others cost $5 each. The total cost is $70. Let x be the number of $3 notebooks and y the number of $5 notebooks. Then x + y = 18 and 3x + 5y = 70.
Subtract three times the first equation from the second: 2y = 16, so y = 8 and x = 10. Check: 10 + 8 = 18 and 3(10) + 5(8) = 70. The context also requires nonnegative whole-number counts, which these values satisfy.
Worked example: concentration has its own equation
Mix x litres of a 20% solution with y litres of a 50% solution to make 10 litres of a 32% solution. Assume the solution volumes add. The volume equation is x + y = 10. The component equation is 0.20x + 0.50y = 3.20, because 32% of 10 litres is 3.20 litres.
Subtract 0.20 times the first equation: 0.30y = 1.20, hence y = 4 and x = 6. Check the component amount: 1.20 + 2.00 = 3.20 litres. Writing 20x + 50y = 32 would mix percentage values with a total concentration and omit the total volume.
Choose elimination or substitution for the actual target
Use substitution when a variable is already isolated or easy to isolate. Use elimination when coefficients already align or can align with a simple multiplication. You may not need both variables if the question asks for an expression.
For 2x + y = 7 and x + 2y = 8, adding gives 3x + 3y = 15, so x + y = 5. That answers an x + y question directly. Solving separately gives x = 2 and y = 3, which confirms the result but is extra work.
One solution, no solution, or infinitely many
The equations 2x + 3y = 12 and 4x + 6y = 24 describe the same line and have infinitely many solution pairs. Changing the second constant to 25 makes the lines parallel and inconsistent: doubling the first equation requires 24, not 25. If the coefficients are not proportional, the lines intersect once.
For a parameter example, compare 2x + 3y = 12 with 4x + ky = 24. The system has infinitely many solutions when k = 6. For k ≠ 6 it has one solution. A different right-hand side would change the conclusion, so compare constants as well as variable coefficients.
Verify the result against both conditions
Substitute the pair into both original equations and answer the quantity the question requests. In context, check units and whether negative or fractional values make sense. For a graph, check whether its axes use the same units as the equations.
Use your practice log to distinguish translation, arithmetic, solution-count, and final-answer errors. For a system containing a quadratic or other nonlinear relationship, continue to nonlinear equations and systems.
Official sources
Next step