Solve systems of linear equations by substitution and elimination, recognize systems with one, no, or infinitely many solutions, and find the solution region of a system of inequalities.

What this topic is asking

The Algebra category requires you to solve systems of linear equations (the A-REI standards) by substitution and elimination, to know how many solutions a system can have, and to find the solution region of a system of inequalities. On the Grade 10 MCAS, systems appear as selected-response items and frequently as a constructed-response word problem, where defining variables and showing the system are worth explicit credit.

Substitution

Substitution works best when one equation already gives a variable alone. Replace that variable in the other equation, solve the resulting single-variable equation, then back-substitute.

For $y = 3x - 4$ and $x + y = 8$: substitute to get $x + (3x - 4) = 8$, so $4x - 4 = 8$, $x = 3$, and $y = 3(3) - 4 = 5$. The solution is $(3, 5)$. Always state the answer as an ordered pair and check it in the equation you did not use to solve.

Elimination

Elimination adds the two equations so that one variable cancels. If neither variable cancels directly, scale one or both equations first so a pair of coefficients are opposites.

For $2x + 3y = 12$ and $2x - y = 4$: subtract the equations (or add after negating the second) so the $2x$ terms cancel: $(2x + 3y) - (2x - y) = 12 - 4$, giving $4y = 8$, so $y = 2$. Back-substitute: $2x - 2 = 4$, so $x = 3$. The solution is $(3, 2)$.

How many solutions

The number of solutions reflects how the lines relate:

• One solution: the lines have different slopes and cross at a single point.
• No solution: the lines are parallel, with the same slope but different intercepts. Algebraically, the variables cancel and you get a false statement like $0 = 5$.
• Infinitely many: the equations describe the same line. Algebraically, the variables cancel and you get a true statement like $0 = 0$.

So if solving a system collapses to $0 = 7$, there is no solution; if it collapses to $0 = 0$, there are infinitely many.

Solving by graphing

A system can also be solved by graphing both lines and reading the intersection point. This is the method behind many technology-enhanced MCAS items, where you plot two lines on a grid and the crossing point is the solution. Graphing is exact only when the intersection lands on grid points; otherwise it gives an estimate that substitution or elimination then confirms. The graph also makes the three solution cases visible: crossing lines meet once, parallel lines never meet, and coincident lines overlap everywhere.

Linear-quadratic systems

The MCAS also pairs a line with a parabola. Solve a linear-quadratic system by substituting the line into the quadratic, which produces a single quadratic equation with up to two solutions. For $y = x + 2$ and $y = x^2$, substitute to get $x^2 = x + 2$, so $x^2 - x - 2 = 0$, factoring to $(x - 2)(x + 1) = 0$, giving $x = 2$ and $x = -1$. Each $x$ gives a $y$ from the line: $(2, 4)$ and $(-1, 1)$. Graphically these are the two points where the line cuts the parabola, and a line can meet a parabola at two points, one point (tangent), or none.

Systems of inequalities

Graph each inequality as a half-plane: draw the boundary line (solid for $\leq$ or $\geq$, dashed for $<$ or $>$) and shade the side that satisfies it (test a point such as the origin). The solution region is where the shadings overlap. A point is a solution only if it satisfies every inequality, so checking a candidate means substituting into all of them. A constructed-response item may ask you to name a point in the solution region or to verify whether a given point lies in it, which is just a substitution check against each inequality.

Try this

Q1. Solve $y = x + 1$ and $2x + y = 7$.

• Cue. $2x + (x + 1) = 7$, so $x = 2$, $y = 3$: the point $(2, 3)$.

Q2. How many solutions does $y = 4x + 2$ and $y = 4x - 5$ have?

• Cue. Same slope, different intercepts: parallel, so no solution.