Algebra 1 · Unit 4
Check possible solutions with algebra, a graph, or a table, and interpret them in a real-life story.
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Substitute into every piece
A pair (x, y) is a solution of a system of equations only if both equations are true. For inequalities, both inequalities must be true.
A graph is a check
On two lines, the solution is the intersection. On a half-plane or overlap, the point must sit in the shaded region. On a dashed boundary, a point on the line does not count.
A table is a technology check
List x and the two outputs. The solution of an equation system is where the two y-values match.
The story can reject a true pair
Counts, time, and length cannot be negative in most stories. After the algebra checks out, interpret the answer.
Yes or no. Substitute, or read a short table.
3 + 2 = 5, true.
2(3) − 2 = 6 − 2 = 4, true.
Both equations are true, so yes.
Answer: yes
1 + 4 = 5, true.
3(1) − 4 = 3 − 4 = −1.
−1 ≠ 1, so no.
Answer: no
5 ≥ 2 + 1
5 ≥ 3, true.
Yes. The point is above the line y = x + 1.
Answer: yes
0 > 0 + 2 is false.
0 + 0 ≤ 4 is true.
It fails the first inequality, so no.
Answer: no
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| x + 1 | 2 | 3 | 4 | 5 |
| −1x + 7 | 6 | 5 | 4 | 3 |
Look for matching outputs.
When x = 3, both y-values are 4.
The intersection is (3, 4).
Check: 3 + 1 = 4 and −3 + 7 = 4.
Answer: (3, 4)
2(4) + 3(1) = 8 + 3 = 11
11 ≤ 11 is true.
Yes. The point is on the solid boundary.
Answer: yes
Check a system of inequalities, read a graph, or interpret a story.
3 ≥ 2 − 1
3 ≥ 1, true.
2 + 3 = 5 ≤ 6, true.
Yes. Both inequalities hold.
Answer: yes
First line: 2(2) − 1 = 3.
Second line: −2 + 5 = 3.
Both true, and the point is where the lines cross.
Yes.
Answer: yes
3 adult tickets and 5 child tickets.
3 + 5 = 8 tickets.
9(3) + 5(5) = 27 + 25 = 52 dollars.
Both facts match. The counts are whole numbers, so the story accepts the pair.
Answer: 3 adult, 5 child; \$52
4 > −1(−1) + 2
4 > 1 + 2
4 > 3, true.
−1 ≥ 0 is false.
No. It fails x ≥ 0.
Answer: no
| x | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| 3x + 3 | 9 | 12 | 15 | 18 |
Look for the output 12.
When x = 3, 3x + 3 = 12.
Check: 3(3) + 3 = 12.
Answer: x = 3
1 ≤ 4, true.
1 ≥ 1 + 2
1 ≥ 3 is false.
No.
Answer: no
Boundary points, technology, and stories that reject a mathematically true pair.
| x | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| 2x + 3 | 7 | 9 | 11 | 13 |
| x − 3 | −1 | 0 | 1 | 2 |
We need 2x + y = 10 and x − y = 1 with y = 3.
So 2x + 3 = 10 and x − 3 = 1.
From the second column of the second row, x − 3 = 1 when x = 4.
At x = 4, 2x + 3 = 11, not 10.
The table with y locked at 3 never makes both true at once, so y is not 3.
Solve: add 2x + y = 10 and x − y = 1 to get 3x = 11, x = 113, y = 83.
The table did not hit the solution because y was fixed at 3.
Answer: y is not 3; (113, 83)
2 > 2 is false.
(2, 2) is on y = x.
The inequality is strict, so the dashed line is not included.
No.
Answer: no
113 + 43 = 5, true.
8113 + 543 = 883 + 203 = 1083 = 36, true.
The algebra checks. Ticket counts in a story must be whole numbers.
113 tickets is not possible. The story rejects the pair.
Answer: no; tickets must be whole numbers
(0, 4): 0 ≥ 0, 4 ≥ 0, 4 ≤ 6. Yes.
(2, 2): all true, 2 + 2 = 4 ≤ 6. Yes.
(5, −1): y ≥ 0 fails.
(0, 4) and (2, 2) work. (5, −1) does not.
Answer: (0, 4) and (2, 2)
6 liters of 10% and 6 liters of 30%.
6 + 6 = 12 liters total.
0.10(6) + 0.30(6) = 0.6 + 1.8 = 2.4 liters of acid.
Both amounts are positive, so the mix makes sense.
Answer: 6 L of each; 2.4 L acid
≤ includes the boundary, so the line is solid.
Every point on the line makes y = −1x + 4, which satisfies y ≤ −1x + 4.
The point is a solution.
Answer: it is a solution; ≤ includes the line
(1, 3): 1 + 6 = 7 ≤ 8 and 3 + 3 = 6 ≤ 9. Yes.
(3, 1): 3 + 2 = 5 ≤ 8 and 9 + 1 = 10 ≤ 9 is false. No.
(4, 1): 4 + 2 = 6 ≤ 8 and 12 + 1 = 13 ≤ 9 is false. No.
Only (1, 3) is feasible. (4, 1) would be 4 hours of tutoring and 1 hour of group class, which breaks the 3x + y ≤ 9 cap.
Answer: only (1, 3)
Algebra: 3(4) + 2(−1) = 12 − 2 = 10.
4 − 4(−1) = 4 + 4 = 8.
Graph: both lines must pass through (4, −1). If you plotted them, that is the intersection.
A table of both y-values would match at x = 4.
Answer: both equations true; intersection point
(2, 2): 2 > 2 is false and 2 < 2 is false.
Both inequalities are strict, so the line y = x is not included in either.
No point can be both greater than x and less than x.
The system has no solution.
Answer: no solution
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