Solve rational equations to uncover a hidden picture in this engaging math worksheet.

Since the image you uploaded contains a worksheet titled "Solving Rational Expressions Hidden Picture", and there is no visible grid or hidden picture in your description, I will solve each of the 12 rational equations step by step. Once you have the solutions, you can locate them on the grid (which is not included here) and shade the corresponding squares to reveal the hidden picture.

Let’s go through each problem carefully:

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1.

$$
\frac{4}{x+2} + \frac{3}{x+5} = \frac{5}{x^2 + 7x + 10}
$$

Step 1: Factor denominator on right:
$$
x^2 + 7x + 10 = (x+2)(x+5)
$$

So the equation becomes:
$$
\frac{4}{x+2} + \frac{3}{x+5} = \frac{5}{(x+2)(x+5)}
$$

Step 2: Multiply both sides by the LCD: $(x+2)(x+5)$

$$
4(x+5) + 3(x+2) = 5
$$

$$
4x + 20 + 3x + 6 = 5
$$

$$
7x + 26 = 5 \Rightarrow 7x = -21 \Rightarrow x = -3
$$

✔ Solution: $x = -3$

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2.

$$
y + \frac{6}{y} = 5
$$

Multiply both sides by $y$:
$$
y^2 + 6 = 5y
\Rightarrow y^2 - 5y + 6 = 0
\Rightarrow (y-2)(y-3) = 0
$$

✔ Solutions: $y = 2$, $y = 3$

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3.

$$
\frac{5}{x+2} + \frac{x}{x-2} = \frac{8}{x^2 - 4}
$$

Note: $x^2 - 4 = (x+2)(x-2)$

Multiply both sides by $(x+2)(x-2)$:

$$
5(x-2) + x(x+2) = 8
$$

$$
5x - 10 + x^2 + 2x = 8
\Rightarrow x^2 + 7x - 10 = 8
\Rightarrow x^2 + 7x - 18 = 0
\Rightarrow (x+9)(x-2) = 0
$$

So $x = -9$ or $x = 2$

But check for extraneous solutions:
- $x = 2$: makes denominator $x-2 = 0$ → invalid
- $x = -9$: valid

✔ Solution: $x = -9$

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4.

$$
\frac{m+2}{m-1} + \frac{4}{m-5} = \frac{6}{m^2 - 6m + 5}
$$

Factor denominator: $m^2 - 6m + 5 = (m-1)(m-5)$

LCD: $(m-1)(m-5)$

Multiply both sides:

$$
(m+2)(m-5) + 4(m-1) = 6
$$

$$
(m^2 - 5m + 2m - 10) + 4m - 4 = 6
\Rightarrow m^2 - 3m - 10 + 4m - 4 = 6
\Rightarrow m^2 + m - 14 = 6
\Rightarrow m^2 + m - 20 = 0
\Rightarrow (m+5)(m-4) = 0
$$

So $m = -5$ or $m = 4$

Check restrictions: $m \ne 1, 5$ → both valid

✔ Solutions: $m = -5$, $m = 4$

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5.

$$
\frac{7}{s+4} - \frac{2}{s-3} = \frac{2s-9}{s^2 + s - 12}
$$

Factor denominator: $s^2 + s - 12 = (s+4)(s-3)$

LCD: $(s+4)(s-3)$

Multiply both sides:

$$
7(s-3) - 2(s+4) = 2s - 9
$$

$$
7s - 21 - 2s - 8 = 2s - 9
\Rightarrow 5s - 29 = 2s - 9
\Rightarrow 3s = 20 \Rightarrow s = \frac{20}{3}
$$

✔ Solution: $s = \frac{20}{3}$

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6.

$$
\frac{a}{a+5} = \frac{3}{a+1}
$$

Cross-multiply:

$$
a(a+1) = 3(a+5)
\Rightarrow a^2 + a = 3a + 15
\Rightarrow a^2 - 2a - 15 = 0
\Rightarrow (a-5)(a+3) = 0
$$

So $a = 5$ or $a = -3$

Check restrictions: $a \ne -5, -1$ → both valid

✔ Solutions: $a = 5$, $a = -3$

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7.

$$
\frac{3}{m-1} = \frac{2m}{m+4}
$$

Cross-multiply:

$$
3(m+4) = 2m(m-1)
\Rightarrow 3m + 12 = 2m^2 - 2m
\Rightarrow 0 = 2m^2 - 5m - 12
\Rightarrow (2m + 3)(m - 4) = 0
$$

So $m = -\frac{3}{2}$ or $m = 4$

Check restrictions: $m \ne 1, -4$ → both valid

✔ Solutions: $m = -\frac{3}{2}, 4$

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8.

$$
2 = \frac{y}{y+3} - \frac{3}{y-5}
$$

LCD: $(y+3)(y-5)$

Multiply both sides:

$$
2(y+3)(y-5) = y(y-5) - 3(y+3)
$$

Left side: $2(y^2 - 2y - 15) = 2y^2 - 4y - 30$

Right side: $y^2 - 5y - 3y - 9 = y^2 - 8y - 9$

Now:
$$
2y^2 - 4y - 30 = y^2 - 8y - 9
\Rightarrow y^2 + 4y - 21 = 0
\Rightarrow (y+7)(y-3) = 0
$$

So $y = -7$ or $y = 3$

Check restrictions: $y \ne -3, 5$ → both valid

✔ Solutions: $y = -7$, $y = 3$

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9.

$$
\frac{4u}{u+3} + u = \frac{8}{u+3}
$$

Multiply both sides by $u+3$:

$$
4u + u(u+3) = 8
\Rightarrow 4u + u^2 + 3u = 8
\Rightarrow u^2 + 7u - 8 = 0
\Rightarrow (u+8)(u-1) = 0
$$

So $u = -8$ or $u = 1$

Check restriction: $u \ne -3$ → both valid

✔ Solutions: $u = -8$, $u = 1$

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10.

$$
\frac{2}{w} + 3 = \frac{7}{w+6}
$$

Multiply both sides by $w(w+6)$:

$$
2(w+6) + 3w(w+6) = 7w
$$

$$
2w + 12 + 3w^2 + 18w = 7w
\Rightarrow 3w^2 + 20w + 12 = 7w
\Rightarrow 3w^2 + 13w + 12 = 0
\Rightarrow (3w + 4)(w + 3) = 0
$$

So $w = -\frac{4}{3}$ or $w = -3$

Check restrictions: $w \ne 0, -6$ → both valid

✔ Solutions: $w = -\frac{4}{3}, -3$

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11.

$$
\frac{1}{x+3} + \frac{5}{x^2 - 9} = \frac{2}{x-3}
$$

Note: $x^2 - 9 = (x+3)(x-3)$

LCD: $(x+3)(x-3)$

Multiply:

$$
1(x-3) + 5 = 2(x+3)
\Rightarrow x - 3 + 5 = 2x + 6
\Rightarrow x + 2 = 2x + 6
\Rightarrow -x = 4 \Rightarrow x = -4
$$

Check: $x \ne -3, 3$ → valid

✔ Solution: $x = -4$

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12.

$$
\frac{-1}{y-3} = \frac{7y+3}{y^2 - 8y + 15} + \frac{3y}{y-5}
$$

Factor: $y^2 - 8y + 15 = (y-3)(y-5)$

LCD: $(y-3)(y-5)$

Multiply both sides:

$$
-1(y-5) = (7y+3) + 3y(y-3)
$$

Left: $-y + 5$

Right: $7y + 3 + 3y^2 - 9y = 3y^2 - 2y + 3$

So:
$$
-y + 5 = 3y^2 - 2y + 3
\Rightarrow 0 = 3y^2 - y - 2
\Rightarrow (3y + 2)(y - 1) = 0
$$

So $y = -\frac{2}{3}$ or $y = 1$

Check restrictions: $y \ne 3, 5$ → both valid

✔ Solutions: $y = -\frac{2}{3}, 1$

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✔ Final Answers Summary:

| Problem | Solutions |
|--------|-----------|
| 1 | $x = -3$ |
| 2 | $y = 2, 3$ |
| 3 | $x = -9$ |
| 4 | $m = -5, 4$ |
| 5 | $s = \frac{20}{3}$ |
| 6 | $a = 5, -3$ |
| 7 | $m = -\frac{3}{2}, 4$ |
| 8 | $y = -7, 3$ |
| 9 | $u = -8, 1$ |
| 10 | $w = -\frac{4}{3}, -3$ |
| 11 | $x = -4$ |
| 12 | $y = -\frac{2}{3}, 1$ |

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🔍 Instructions for the Grid:

Once you have this worksheet printed, look at the grid (not shown here). Each square likely has a number or coordinate. For each problem, find the solution(s) on the grid and shade the square(s) that contain those values.

For example:
- If $x = -3$ appears on the grid, shade that square.
- If $y = 2$ and $y = 3$ appear, shade both.

After shading all correct squares based on the solutions above, a hidden picture should emerge!

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Let me know if you’d like help interpreting the grid or creating a visual version of the shaded results!