o Substitution
-> Solve one equation for one variable.
-> Substitute that expression into the other equation.
-> Solve the resulting one-variable equation.
-> Back-substitute to find the other variable.
-> Example: x + 3y = 7 and 2x - y = 0.
-> From the second equation, y = 2x.
-> Substitute: x + 3(2x) = 7.
-> Then 7x = 7, so x = 1 and y = 2.
-> Check: 1 + 3(2) = 7 and 2(1) - 2 = 0.
o Elimination
-> Line up like terms in both equations.
-> Add or subtract to remove one variable.
-> Example: 3x + 4y = 11 and 3x - 2y = 5.
-> Subtract: 6y = 6, so y = 1.
-> Substitute into the first: 3x + 4 = 11.
-> Then 3x = 7 and x = 7/3.
-> The solution is (7/3, 1).
-> Check the second: 3(7/3) - 2 = 7 - 2 = 5.
-> Opposite terms add away; equal terms subtract away.
o Scaling
-> Multiply equations by nonzero constants to match terms.
-> Example: 2x + 3y = 8 and 5x - 2y = 1.
-> Multiply by 2 and 3: 4x + 6y = 16.
-> The second becomes 15x - 6y = 3.
-> Add: 19x = 19, so x = 1.
-> Then 2(1) + 3y = 8 gives y = 2.
-> Check: 5(1) - 2(2) = 1.
-> Scaling changes the form, not the solution set.
o Special Outcomes
-> Example: 2x + 4y = 6 and x + 2y = 3.
-> The second is the first divided by 2.
-> Same line: infinitely many solutions.
-> Multiply the second by -2 and add: 0 = 0.
-> Example: 2x + 4y = 6 and x + 2y = 4.
-> Proportional left sides, different constants.
-> Multiply the second by -2 and add: 0 = -2.
-> This is false, so there is no solution.
-> A true identity is dependent; a contradiction is inconsistent.
o Mixture Model
-> Track total amount and the active ingredient.
-> Example: blend 12 L of 20% juice with 8 L of 45% juice.
-> Total 20 L; juice: 0.20(12) + 0.45(8) = 6 L.
-> Concentration: 6/20 = 0.30, or 30%.
-> Reverse problem: how much 45% juice makes 20 L at 30%?
-> Let x be liters of 45% juice; then 20 - x is 20% juice.
-> 0.45x + 0.20(20 - x) = 6.
-> 0.25x + 4 = 6 gives x = 8 L of 45% juice.
-> Then 12 L of 20% juice; check 0.45(8) + 0.20(12) = 6.
o Checks
-> Substitute into both original equations; one is not enough.
-> Check units and whether the answer is viable.
-> Example: (3, -1) in 2x + y = 5 and x - y = 4.
-> First: 2(3) - 1 = 5, true.
-> Second: 3 - (-1) = 4, true.
-> Negative amounts can satisfy algebra but fail context.
-> For whole items, check that an integer solution exists.
-> Report the solution with the variable meanings.