Math worksheet for solving systems of linear and quadratic equations by graphing, featuring three problems with graph paper grids.

Let’s solve each system step by step. We’re solving systems of equations by graphing — that means we’ll find where the two graphs (one linear, one quadratic) intersect. Those intersection points are the solutions.

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Problem 1:

Equations:
- y = x + 3 → This is a straight line.
- y = x² - 8x + 15 → This is a parabola.

To solve by graphing, we can also solve algebraically to check our work (since we don’t have actual graph paper here). Set them equal:

x + 3 = x² - 8x + 15

Bring all terms to one side:

0 = x² - 8x + 15 - x - 3
→ 0 = x² - 9x + 12

Wait — let me double-check that subtraction:

x² - 8x + 15 minus (x + 3) = x² - 8x + 15 - x - 3 = x² - 9x + 12 ✔️

Now solve x² - 9x + 12 = 0

Use quadratic formula:

x = [9 ± √(81 - 48)] / 2 = [9 ± √33] / 2

√33 ≈ 5.744

So:

x₁ ≈ (9 + 5.744)/2 ≈ 14.744/2 ≈ 7.372
x₂ ≈ (9 - 5.744)/2 ≈ 3.256/2 ≈ 1.628

Then plug into y = x + 3:

For x ≈ 7.372 → y ≈ 10.372
For x ≈ 1.628 → y ≈ 4.628

But wait — maybe I made an error? Let me recheck the original equations.

Original problem says:

y = x + 3
y = x² - 8x + 15

Set equal:

x + 3 = x² - 8x + 15
→ 0 = x² - 9x + 12 → correct.

Discriminant: b² - 4ac = 81 - 48 = 33 → not a perfect square → irrational roots.

But perhaps the problem expects exact form? Or maybe I misread the equation?

Wait — looking back at the image description in the user's upload (even though I’m not supposed to mention it), the first problem might be:

Actually, let me re-express carefully.

Perhaps there was a typo in my reading? Let me try factoring x² - 9x + 12 — doesn't factor nicely.

Alternatively, maybe the quadratic is different? Let me assume the problem is as written.

But for graphing purposes, we’d plot both and see intersections. Since this is homework, likely the numbers are chosen to give nice answers. Did I copy wrong?

Wait — let me check again: User wrote:

"1
y = x + 3
y = x² - 8x + 15"

Yes.

But x² - 8x + 15 factors as (x-3)(x-5). So vertex at x=4, y=16-32+15=-1.

Line y=x+3 has slope 1, y-intercept 3.

At x=3: line y=6, parabola y=9-24+15=0 → not equal
At x=5: line y=8, parabola y=25-40+15=0 → not equal
At x=4: line y=7, parabola y=16-32+15=-1 → no

Try x=1: line y=4, parabola y=1-8+15=8 → no
x=2: line y=5, parabola y=4-16+15=3 → no
x=6: line y=9, parabola y=36-48+15=3 → no
x=7: line y=10, parabola y=49-56+15=8 → no
x=8: line y=11, parabola y=64-64+15=15 → no

No integer solutions. So perhaps the problem is designed to use calculator or approximate? But that seems odd for a worksheet titled “by graphing” with grids.

Wait — maybe I misread the quadratic? What if it’s y = x² - 8x + 12? Then:

x + 3 = x² - 8x + 12 → 0 = x² -9x +9 → still not nice.

Or y = x² - 8x + 16? Then:

x+3 = x²-8x+16 → 0=x²-9x+13 → discriminant 81-52=29 → no.

Wait — what if the line is y = x + 5? Then:

x+5 = x²-8x+15 → 0=x²-9x+10 → (x-2)(x-7)=0 → x=2 or 7 → y=7 or 12 → nice!

But the problem says y = x + 3.

Alternatively, maybe the quadratic is y = x² - 6x + 8? Let’s test:

x+3 = x²-6x+8 → 0=x²-7x+5 → discriminant 49-20=29 → no.

I think I need to proceed with the given equations.

But let me calculate numerically more accurately.

x² - 9x + 12 = 0

x = [9 ± √(81 - 48)] / 2 = [9 ± √33]/2

√33 = sqrt(33) ≈ 5.7445626465

So:

x1 = (9 + 5.7445626465)/2 = 14.7445626465/2 = 7.37228132325
y1 = x1 + 3 = 10.37228132325

x2 = (9 - 5.7445626465)/2 = 3.2554373535/2 = 1.62771867675
y2 = x2 + 3 = 4.62771867675

So solutions approximately (1.63, 4.63) and (7.37, 10.37)

But since the grid is probably for integer coordinates, maybe the problem has a typo? Or perhaps I should leave it exact?

The instruction says "solve by graphing", so on graph paper, you'd plot and estimate.

But for final answer, perhaps they want exact form?

Let me move to Problem 2 and 3 to see if they are nicer.

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Problem 2:

Equations:
- y = x + 3
- y = -x² + 2x + 3

Set equal:

x + 3 = -x² + 2x + 3

Subtract x + 3 from both sides:

0 = -x² + 2x + 3 - x - 3
→ 0 = -x² + x
→ 0 = x(-x + 1)
→ x = 0 or x = 1

Check:

If x=0: y = 0+3=3; and y = -0 +0 +3=3 → good
If x=1: y=1+3=4; y=-1+2+3=4 → good

Perfect! Solutions: (0,3) and (1,4)

This makes sense for graphing — easy to plot.

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Problem 3:

Equations:
- y = -x + 6
- y = x² + 8x + 6

Set equal:

-x + 6 = x² + 8x + 6

Subtract (-x + 6) from both sides:

0 = x² + 8x + 6 + x - 6
→ 0 = x² + 9x
→ 0 = x(x + 9)

So x = 0 or x = -9

Find y:

If x=0: y = -0 + 6 = 6; and y = 0 + 0 + 6 = 6 → good
If x=-9: y = -(-9) + 6 = 9+6=15; and y = 81 -72 +6 = 15 → good

Solutions: (0,6) and (-9,15)

Also nice integers.

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Back to Problem 1: Why isn’t it nice? Maybe I misread the quadratic.

Looking back at the user’s input: “y = x² - 8x + 15”

But 15 is 3*5, and -8x suggests vertex at x=4.

Perhaps the line is y = x - 3? Let’s try:

x - 3 = x² -8x +15 → 0 = x² -9x +18 → (x-3)(x-6)=0 → x=3,y=0; x=6,y=3 → nice!

But the problem says y = x + 3.

Another possibility: maybe it’s y = x² - 8x + 12? Then:

x+3 = x²-8x+12 → 0=x²-9x+9 → still not integer.

Or y = x² - 8x + 16? Then x+3 = x²-8x+16 → 0=x²-9x+13 → no.

Wait — what if the quadratic is y = x² - 6x + 8? Then:

x+3 = x²-6x+8 → 0=x²-7x+5 → no.

Perhaps it’s y = x² - 8x + 15, and we accept irrational solutions.

But for a school worksheet, unlikely.

Another idea: maybe the first equation is y = -x + 3? Let’s try:

- x + 3 = x² -8x +15 → 0 = x² -7x +12 → (x-3)(x-4)=0 → x=3,y=0; x=4,y=-1 → but y=-x+3 at x=3 is 0, at x=4 is -1 → matches.

But the problem says y = x + 3, not -x + 3.

I think there might be a typo in the problem, but since I have to go with what’s given, I’ll provide the exact solutions for Problem 1.

However, let me double-check the user’s text: “y = x + 3” and “y = x² - 8x + 15”

Perhaps in the image, it’s different, but I must rely on the text provided.

Maybe for Problem 1, they expect us to graph and estimate, but for final answer, we can write exact values.

So for Problem 1:

Solutions are x = [9 ± √33]/2, y = x + 3

But typically, we write as ordered pairs.

Since Problems 2 and 3 have integer solutions, and Problem 1 does not, perhaps I made a mistake in setting up.

Wait — let me solve Problem 1 again carefully.

Given:
Eq1: y = x + 3
Eq2: y = x² - 8x + 15

Set equal:
x + 3 = x² - 8x + 15

Move all to right:
0 = x² - 8x + 15 - x - 3
0 = x² - 9x + 12

Yes.

Discriminant D = 81 - 48 = 33

Roots: [9 ± √33]/2

√33 cannot be simplified.

So solutions are:

( \frac{9 - \sqrt{33}}{2}, \frac{15 - \sqrt{33}}{2} ) and ( \frac{9 + \sqrt{33}}{2}, \frac{15 + \sqrt{33}}{2} )

Because y = x + 3, so if x = [9±√33]/2, then y = [9±√33]/2 + 3 = [9±√33 + 6]/2 = [15±√33]/2

Yes.

But for a student, perhaps they want decimal approximations.

Since the other problems have integer answers, and this one doesn't, I suspect a typo, but I'll proceed.

Perhaps the quadratic is y = x² - 8x + 12? Let me calculate that quickly.

If y = x² - 8x + 12, then:

x+3 = x²-8x+12 → 0=x²-9x+9 → x=[9±√(81-36)]/2=[9±√45]/2=[9±3√5]/2 — still not nice.

Or y = x² - 8x + 16: x+3 = x²-8x+16 → 0=x²-9x+13 — no.

Another possibility: maybe the line is y = 2x + 3? Then:

2x+3 = x²-8x+15 → 0=x²-10x+12 — discriminant 100-48=52 — no.

I think I have to accept that Problem 1 has irrational solutions.

But let's look at the grid size — probably 10x10 or something, so for x from -5 to 5 or so, but our solutions are at x≈1.6 and 7.4, which might be on the grid if it goes to 10.

In many worksheets, grids go from -10 to 10.

So perhaps it's fine.

For the sake of completing, I'll provide exact forms for Problem 1, and integer for others.

But the instruction is to solve by graphing, so perhaps they expect the student to plot and read off approximate values.

However, for final answer, I'll give exact where possible.

Let me summarize:

Problem 1:
Solutions:
x = \frac{9 \pm \sqrt{33}}{2}, y = \frac{15 \pm \sqrt{33}}{2}

Or approximately (1.63, 4.63) and (7.37, 10.37)

Problem 2:
Solutions: (0,3) and (1,4)

Problem 3:
Solutions: (0,6) and (-9,15)

Now, for the final answer, since the user didn't specify format, I'll list all solutions.

But the worksheet has three separate problems, so I should present each.

Also, in the response, I need to follow the structure.

Since ExplainAnswer is false, TeachTopic is false, GeneratePractice is false, I only provide Final Answer.

But the rules say: "Complete ALL reasoning and explanations BEFORE writing the Final Answer." and "Final Answer: [Provide only the final correct answer]"

And "If an option is false, completely omit that section."

So I will do the reasoning above, then in Final Answer, provide the solutions.

But how to format? Probably list for each problem.

Also, for Problem 1, since it's messy, perhaps I should box the exact form.

Maybe the problem is intended to be solved by graphing, so approximate is acceptable, but for accuracy, I'll give exact.

Another thought: in Problem 1, if I factor the quadratic: x² -8x+15 = (x-3)(x-5), so roots at 3 and 5, vertex at (4,-1)

Line y=x+3 passes through (0,3), (1,4), etc.

At x=3, parabola y=0, line y=6 — difference 6
At x=4, parabola y=16-32+15= -1, line y=7 — difference 8
At x=5, parabola y=0, line y=8 — difference 8
At x=2, parabola y=4-16+15=3, line y=5 — difference 2
At x=1, parabola y=1-8+15=8, line y=4 — difference 4
At x=6, parabola y=36-48+15=3, line y=9 — difference 6
At x=7, parabola y=49-56+15=8, line y=10 — difference 2
At x=8, parabola y=64-64+15=15, line y=11 — difference 4

So between x=1 and 2, and between x=7 and 8, they cross, as we calculated.

So no integer solutions.

I think for the final answer, I'll provide:

For Problem 1: \left( \frac{9 - \sqrt{33}}{2}, \frac{15 - \sqrt{33}}{2} \right) and \left( \frac{9 + \sqrt{33}}{2}, \frac{15 + \sqrt{33}}{2} \right)

For Problem 2: (0, 3) and (1, 4)

For Problem 3: (0, 6) and (-9, 15)

To make it neat, perhaps write as:

Solution(s) for 1: \left( \frac{9 \pm \sqrt{33}}{2}, \frac{15 \pm \sqrt{33}}{2} \right)

But usually we list both.

Since the worksheet has "Solution(s): ________", probably expect ordered pairs.

For Problem 2 and 3, it's clear.

For Problem 1, perhaps they want decimal, but I think exact is better.

I recall that in some curricula, they leave it in radical form.

So I'll go with that.

Final Answer:

Problem 1: \left( \frac{9 - \sqrt{33}}{2}, \frac{15 - \sqrt{33}}{2} \right) and \left( \frac{9 + \sqrt{33}}{2}, \frac{15 + \sqrt{33}}{2} \right)

Problem 2: (0, 3) and (1, 4)

Problem 3: (0, 6) and (-9, 15)

To write it compactly.

Since the user might expect a specific format, but I think this is fine.

Perhaps for Problem 1, simplify the y-coordinate.

As I did earlier, y = x + 3, so if x = [9±√33]/2, y = [9±√33]/2 + 6/2 = [15±√33]/2, yes.

So it's correct.

Now for the Final Answer section.