Algebra I: Systems & Linear Inequalities

How to do it:

1. Graph both lines and read the intersection, or

2. Isolate one variable and substitute into the other equation, or

3. Multiply one equation so a variable cancels when you add.

4. Plug the pair back into both originals.

Worked example: $x+y=10$ and $x-y=2\text{.}$ Add: $2x=12\text{,}$ $x=6\text{,}$ then $y=4\text{.}$ Check: $6+4=10$ and $6-4=2\text{.}$

Check: parallel distinct lines mean no solution. The same line twice means infinitely many. Grade 8 systems built this; here we keep units and inequality follow-ups. Use the graphing calculator to see the intersection.

How to do it:

1. Graph the boundary $ax+by=c$ (solid for $\leq$ or $\geq\text{,}$ dashed for $<$ or $>\text{).}$

2. Test a point not on the line, often $(0,0)$ if it is not on the boundary.

3. Shade the side that makes the inequality true.

Worked example: $y > 2x + 1\text{.}$ Dashed line through $(0,1)$ with slope 2. $(0,0)\text{:}$ $0 > 1$ is false, so shade the other side.

Check: shading the wrong side is the most common error. Always test. A solid line includes the boundary; a dashed line does not.

How to do it:

1. Graph each inequality as its own half-plane.

2. The solution is the overlap.

3. In a modeling story, drop points that fail a context constraint (negative tickets, over budget).

Worked example: $x \geq 0\text{,}$ $y \geq 0\text{,}$ $x+y \leq 10$ is a triangle in the first quadrant including the axes. A point $(6,5)$ is outside because $11 > 10\text{.}$

Check: a point on a dashed edge is not a solution. A point on a solid edge is, if it also satisfies the other inequalities.

How to do it:

1. Pick the equation where a variable is already alone (or easiest to isolate), and solve it for that variable.

2. Substitute that expression into the other equation, so it has one variable.

3. Solve, then back-substitute to find the second variable, and check both originals.

Worked example: $y=2x-1$ and $3x+y=14\text{.}$ Replace $y$ in the second: $3x+(2x-1)=14\text{,}$ so $5x=15$ and $x=3\text{.}$ Then $y=2(3)-1=5\text{.}$ The solution is $(3,5)\text{.}$

Check: $3(3)+5=14$ and $5=2(3)-1\text{.}$ Substituting into the same equation you solved just gives $0=0$ and tells you nothing.

How to do it:

1. Line up the like terms. Pick the variable to eliminate.

2. Multiply one or both equations by numbers that make that variable's coefficients opposites.

3. Add the equations, solve, and back-substitute.

Worked example: $2x+3y=12$ and $3x-2y=5\text{.}$ To eliminate $y\text{,}$ multiply the first by $2$ and the second by $3\text{:}$ $4x+6y=24$ and $9x-6y=15\text{.}$ Add: $13x=39\text{,}$ so $x=3\text{.}$ Then $2(3)+3y=12$ gives $y=2\text{.}$ Solution $(3,2)\text{.}$

Check: $3(3)-2(2)=5\text{.}$ Multiply the whole equation, right side included; scaling only the left side changes the problem.

How to do it:

1. Write both equations in slope-intercept form and compare.

2. Different slopes: the lines cross once, one solution.

3. Same slope, different intercepts: parallel lines, no solution. Same slope and same intercept: the same line, infinitely many.

4. In elimination, a false statement like $0=4$ means none; a true one like $0=0$ means infinitely many.

Worked example: $y=2x+1$ and $4x-2y=6\text{.}$ The second becomes $y=2x-3\text{:}$ same slope $2\text{,}$ different intercepts, so no solution. By substitution, $4x-2(2x+1)=6$ gives $-2=6\text{,}$ false.

Check: parallel lines never meet, so "no solution" is the honest answer, not $x=0\text{.}$ And $y=2x+1$ with $2y=4x+2$ is the same line twice: every point on it is a solution.

How to do it:

1. Name two unknowns with units.

2. Turn each fact in the story into an equation: one about how many, one about how much.

3. Solve, then check that the answer makes sense in context (no negative tickets).

Worked example: $12$ tickets sold for $\$75\text{;}$ adult tickets cost $\$8$ and child tickets $\$5\text{.}$ Let $a$ be adult tickets and $c$ child tickets. Count: $a+c=12\text{.}$ Money: $8a+5c=75\text{.}$ Substitute $c=12-a\text{:}$ $8a+5(12-a)=75\text{,}$ so $3a+60=75$ and $a=5\text{,}$ $c=7\text{.}$

Check: $5+7=12$ tickets and $8(5)+5(7)=40+35=75$ dollars. Both facts, both units.

How to do it:

1. Graph when you want a picture or an approximate answer, or the numbers are messy.

2. Substitute when a variable is already isolated (like $y=2x-1\text{).}$

3. Eliminate when the equations are in $ax+by=c$ form and a coefficient matches or is easy to match.

Worked example: for $y=3x+2$ with $x+y=10\text{,}$ substitute. For $2x+3y=12$ with $3x-2y=5\text{,}$ eliminate. For a question that just asks roughly where two lines meet, graph and read.

Check: every method gives the same solution; the choice changes only how much arithmetic you do. When in doubt, elimination works for any pair of linear equations.