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Solve systems of two linear equations by substitution and elimination, including word problems, and shade the region a two-variable inequality describes.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn the two standard ways of solving a pair of equations at once — substituting one into the other, and combining them so a variable cancels — and how to write a word problem as a system in the first place. Then you move from a line to a region: a two-variable inequality is solved by a whole half-plane, and a system of them by the overlap.
You already solve one equation in one unknown, and you already read $y = 3x - 1$ as a rule that gives $y$ from $x$. A system is two of those at once, and the whole method is a way of turning two equations in two unknowns into one equation in one unknown — which is a problem you can already do.
System of equations: two or more equations that must hold at the same time.
Solution of a system: a pair $(x, y)$ that satisfies every equation, not just one.
Substitution: replacing a variable by an expression equal to it.
Elimination: adding or subtracting whole equations so that one variable cancels out.
A solution of a system is a pair that satisfies both equations — on a graph, the point where the two lines cross. Substitution suits a system where one equation already says what a variable is: if $y = 3x - 1$, replace $y$ in the other equation and you are left with one equation in $x$ alone. Elimination suits a system in the form $ax + by = c$: multiply one or both equations so a variable's coefficients match in size, then add or subtract to remove it. Either way, once you have one value you substitute it back to get the other. A word problem is the same work with the writing-down done first: name the two unknowns, write one equation per fact.
Another way: diagram
Two lines drawn on a grid, $y = 3x - 1$ and $2x + y = 9$, crossing at the marked point $(2, 5)$, with dotted lines dropped to the axes.
Another way: story
Both methods do the same thing: get rid of one letter. Substitution replaces it; elimination cancels it. Choose whichever costs less writing for the system in front of you.
"I found $x$, so I am finished." A solution is a pair. Substitute the value you found back into either original equation to get the other.
"Just add the equations." Adding only helps when one variable's coefficients are opposites. $2x + 3y = 7$ and $5x - y = 9$ add to $7x + 2y = 16$, which has removed nothing.
"Every system has one solution." Two lines with the same slope and different intercepts never meet, so the system has none; the same line written twice has infinitely many.
The first equation says what $y$ is, so replace $y$ in the second: $2x + (3x - 1) = 9$.
Substitution.
Collect: $5x - 1 = 9$, so $5x = 10$ and $x = 2$.
Substitute back: $y = 3(2) - 1 = 5$. The solution is $(2, 5)$.
Check it in both equations.
The $y$ coefficients are $+2$ and $-2$, so adding removes $y$: $6x = 24$.
Elimination.
So $x = 4$.
Substitute into the first: $12 + 2y = 16$, so $y = 2$. The solution is $(4, 2)$.
The $y$ terms are $+y$ and $-y$, so add the equations: $3x = 9$.
So $x = 3$, and $3 - y = 2$ gives $y = 1$: the solution is $(3, 1)$.
Solve by substitution: $y = 6x + 6$ and $2x + y = 46$. Fill in both values.
$x = $ x $,\ y = $ y
Solve by elimination: $4x + 2y = -4$ and $4x - 2y = -28$. Fill in both values.
$x = $ x $,\ y = $ y
Adult tickets cost $\$7$ and child tickets $\$6$. $17$ tickets were sold for $\$114$ in total. Fill in the counts.
adults $= $ a $,\ $ children $= $ c
To eliminate $x$ from $3x + 7y = 16$ and $5x - 8y = 17$, what could you do?
A rectangle has a perimeter of $58$ cm, and it is $5$ cm longer than it is wide. How long is it?
answer
You already draw the line $y = 2x + 1$ and you already check whether a point lies on it by substituting. An inequality asks the same question with a looser answer: not 'is the point on the line' but 'which side of it is the point on', and the check is the same substitution.
Half-plane: everything on one side of a line — the solution set of a two-variable linear inequality.
Boundary line: the line you get by replacing the inequality sign with $=$. It is solid for $\le$ and $\ge$, dashed for $<$ and $>$.
Test point: a point substituted into the inequality to decide which side to shade. $(0, 0)$ is the easiest whenever the line does not pass through it.
Constraint: an inequality that comes from a limit in a situation — a budget, a capacity, a minimum.
A linear inequality in two variables is solved by every point on one side of a line. Two decisions make the picture. First the boundary: draw $y = mx + b$, solid if the sign is $\le$ or $\ge$, dashed if it is $<$ or $>$. Then the side: take any point not on the line — $(0, 0)$ if it is available — substitute it, and shade the side it is on if the statement comes out true, the other side if it comes out false. That test is reliable even when the inequality has not been rearranged. For a system of inequalities, shade each one and keep only the overlap: a solution has to satisfy all of them at once, which is exactly what a budget or a capacity constraint means in practice.
Another way: diagram
A grid with the dashed line $y = 2x + 1$ and the region above it shaded, and the test point $(0, 0)$ marked with '$0 > 1$ false' beside it, so the shading is on the other side.
Another way: story
An inequality is a rule about a whole area of the plane, not a curve through it. Testing one point is enough because the line is the only place the truth can change: on one side the statement is true everywhere, on the other it is false everywhere.
"$>$ means shade above, always." Only when the inequality is solved for $y$. $2x - y > 4$ is $y < 2x - 4$, which shades below. Solve for $y$ first, or test a point.
"The line is always dashed." $\le$ and $\ge$ include the boundary, so their lines are solid: the points on them really are solutions.
"The solution of a system is everything shaded." It is everything shaded twice. A point has to satisfy both inequalities, so the answer is the overlap, not the union.
Draw $y = 2x + 1$ dashed, because $>$ does not include the line.
Strict sign, dashed boundary.
Test $(0, 0)$: is $0 > 2(0) + 1$? That says $0 > 1$, which is false.
Pick the easiest point off the line.
So shade the other side — the region above the line.
Substitute: $3 + 2(1) = 5$.
Both coordinates go in.
Is $5 \le 5$? Yes — $\le$ allows equality, so the point is a solution and lies on the boundary.
Substitute both coordinates: the statement becomes $0 < 3$.
That is true, so $(0, 0)$ is a solution and its side is the shaded one.
Which region does $y > 6x - 4$ shade?
Is $(-1, 0)$ a solution of $x + 2y \le 12$?
The boundary line of $y \le 4x + 3$ is drawn...
You may spend at most $\$84$ on pens costing $\$5$ each and books costing $\$9$ each. If you buy $x$ pens and $y$ books, which inequality says so?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Solve by substitution: $y = 2x + 8$ and $4x + y = -22$. Fill in both values.
$x = $ x $,\ y = $ y
What is the solution of the system $y \ge x + 1$ and $y \le 4x + 8$?
You can solve a system by substitution or elimination and shade the region a linear inequality describes. Explain why the solution of a system of two inequalities is the overlap of the shadings rather than everything shaded.
8. Your turn: solve $2x + y = 7$ and $x - y = 2$, step 2
20. Your turn: is $(0, 0)$ a solution of $y < 2x + 3$?, step 2