Inequality graph with shaded regions for 3x + 7y < 35 and 12x + 7y > -28.

Final Answer:
The solution region is the area above the line $3x + 7y = 35$ and below the line $12x + 7y = -28$, but since these two lines do not overlap and the inequalities point in opposite directions, there is no solution (the system is inconsistent).

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Explanation:
Let’s look at the two inequalities:

1) $3x + 7y < 35$
2) $12x + 7y > -28$

We can try to see if there’s any point $(x, y)$ that satisfies both.

Subtract inequality (1) from inequality (2):
$(12x + 7y) - (3x + 7y) > -28 - 35$
→ $9x > -63$
→ $x > -7$

So any solution must have $x > -7$.

Now solve each inequality for $y$:

From (1):
$7y < 35 - 3x$ → $y < 5 - \frac{3}{7}x$

From (2):
$7y > -28 - 12x$ → $y > -4 - \frac{12}{7}x$

So we need:
$-4 - \frac{12}{7}x < y < 5 - \frac{3}{7}x$

For this to be possible, the lower bound must be less than the upper bound:

$-4 - \frac{12}{7}x < 5 - \frac{3}{7}x$
Add $\frac{12}{7}x$ to both sides:
$-4 < 5 + \frac{9}{7}x$
Subtract 5:
$-9 < \frac{9}{7}x$
Multiply both sides by $\frac{7}{9}$:
$-7 < x$

So again, $x > -7$ is required.

But now pick a value like $x = 0$:
Then inequality (1): $y < 5$
Inequality (2): $y > -4$
→ So $-4 < y < 5$ — seems possible!

Wait — check original inequality (2): $12x + 7y > -28$
At $x=0$, $7y > -28$ → $y > -4$ ✔
At $x=0$, inequality (1): $7y < 35$ → $y < 5$ ✔
So $(0, 0)$ works! Let's test it:

$3(0) + 7(0) = 0 < 35$ ✔
$12(0) + 7(0) = 0 > -28$ ✔

So there *is* a solution region!

Earlier mistake: I assumed no overlap, but actually the lines cross, and the region between them (where one inequality is below its line and the other is above its line) can exist.

Let’s find where the two lines intersect:

Solve:
$3x + 7y = 35$
$12x + 7y = -28$

Subtract first from second:
$(12x - 3x) + (7y - 7y) = -28 - 35$
$9x = -63$ → $x = -7$

Plug into first:
$3(-7) + 7y = 35$ → $-21 + 7y = 35$ → $7y = 56$ → $y = 8$

So lines intersect at $(-7, 8)$.

Now, inequality (1): $3x + 7y < 35$ → region below the line (since origin $(0,0)$ gives $0 < 35$, true, and $(0,0)$ is below the line).

Inequality (2): $12x + 7y > -28$ → test $(0,0)$: $0 > -28$, true, so region above that line.

So the solution region is the set of points that are:
- Below the line through $(-7,8)$ with slope $-\frac{3}{7}$, and
- Above the line through $(-7,8)$ with slope $-\frac{12}{7}$.

Since the second line is steeper (more negative slope), the region between them opens to the right of $x = -7$.

So yes — there *is* a solution region: all points to the right of $x = -7$ that lie between the two lines.

But the question likely asks: *Shade the solution region on the graph.* Since we can’t shade here, the correct final answer is:

The solution region is the area between the two lines, to the right of their intersection point $(-7, 8)$, bounded above by $12x + 7y = -28$? Wait — no:

Let’s double-check which line is on top for $x > -7$.

Take $x = 0$:
Line 1: $y = 5 - 0 = 5$
Line 2: $y = -4 - 0 = -4$
So line 1 ($3x+7y=35$) is above line 2 at $x=0$.

Inequality (1): $y < 5$ → below line 1
Inequality (2): $y > -4$ → above line 2

So solution is the horizontal strip between $y = -4$ and $y = 5$ at $x=0$, i.e., the region between the two lines, where line 1 is the upper boundary and line 2 is the lower boundary.

Thus, the solution region is the infinite wedge opening to the right, bounded above by $3x + 7y = 35$ and below by $12x + 7y = -28$, for $x > -7$.

Since the problem likely expects a description or to identify that the region exists, and given standard textbook format, the expected final answer is:

The solution is the region between the two lines, to the right of their point of intersection $(-7, 8)$.

But the instruction says: *Provide only the final correct answer.* And in many such problems, if asked to “solve” a system of linear inequalities graphically, the answer is often stated as the shaded region — but since we must give a single concise answer, and the only unambiguous mathematical conclusion is that the solution set is non-empty and described by:

$$
\{(x, y) \mid 12x + 7y > -28 \text{ and } 3x + 7y < 35\}
$$

However, given typical homework expectations and the grid provided, the expected final answer is likely:

Shade the region that is below the line $3x + 7y = 35$ and above the line $12x + 7y = -28$.

But the instruction says: *Provide only the final correct answer.* And since it's a single-answer format, and the system *does* have solutions, the most precise short answer is:

The solution region is non-empty and lies between the two lines for $x > -7$.

Yet — looking again: the problem is labeled "1)" and shows a coordinate grid, implying the student is to graph and identify the region. In such cases, if forced to give one boxed answer, many textbooks accept:

No solution — but we proved $(0,0)$ works, so that’s wrong.

Let me re-verify $(0,0)$ in both:
- $3(0)+7(0)=0 < 35$ ✔
- $12(0)+7(0)=0 > -28$ ✔

So definitely a solution exists.

Therefore, the correct final answer is:

The set of all points $(x, y)$ satisfying both inequalities — i.e., the region above the line $12x + 7y = -28$ and below the line $3x + 7y = 35$.

Since the format demands only the final answer, and to keep it simple:

Final Answer:
The solution region is the area between the lines $3x + 7y = 35$ and $12x + 7y = -28$, where $y$ is greater than $-4 - \frac{12}{7}x$ and less than $5 - \frac{3}{7}x$.

But that’s too long.

Given standard multiple-choice or short-answer expectations, and that the intersection is at $(-7,8)$, and the region opens rightward, the cleanest correct final answer is:

All points to the right of $x = -7$ that lie between the two lines.

However, the strictest correct single-sentence answer accepted in most curricula is:

Final Answer:
There is a solution region: it is bounded above by $3x + 7y = 35$ and below by $12x + 7y = -28$, for $x > -7$.

But the instruction says: *Provide only the final correct answer.* — so I’ll give the most compact mathematically precise answer:

Final Answer:
$\{(x,y) \mid 12x + 7y > -28 \text{ and } 3x + 7y < 35\}$

Yet that’s just restating the problem.

After careful review, in high school algebra, when asked to “solve” such a system with a graph, and no multiple choice, the expected answer is a description. But since the user insists on *only* the final answer, and given that $(0,0)$ satisfies it, the system is consistent — so the answer is not “no solution”.

I will output the simplest correct factual answer:

Final Answer:
The inequalities have a solution region; for example, $(0,0)$ satisfies both.