Solve each system of equations on the worksheet.

Let's solve each system of equations step by step. We'll use the elimination or substitution method, depending on what’s most efficient.

---

Problem 1:

$$
\begin{aligned}
(1)\quad & \frac{2}{5}x + \frac{3}{4}y = 5 \\
(2)\quad & x - y = 6
\end{aligned}
$$

#### Step 1: Solve equation (2) for $x$:
$$
x = y + 6
$$

#### Step 2: Substitute into equation (1):
$$
\frac{2}{5}(y + 6) + \frac{3}{4}y = 5
$$

Distribute:
$$
\frac{2}{5}y + \frac{12}{5} + \frac{3}{4}y = 5
$$

Combine like terms. First, find a common denominator for $y$ terms (LCM of 5 and 4 is 20):

$$
\left(\frac{8}{20}y + \frac{15}{20}y\right) + \frac{12}{5} = 5 \Rightarrow \frac{23}{20}y + \frac{12}{5} = 5
$$

Convert $\frac{12}{5}$ to twentieths: $\frac{48}{20}$, and $5 = \frac{100}{20}$:

$$
\frac{23}{20}y + \frac{48}{20} = \frac{100}{20}
$$

Subtract:
$$
\frac{23}{20}y = \frac{52}{20} \Rightarrow y = \frac{52}{20} \cdot \frac{20}{23} = \frac{52}{23}
$$

Now plug back into $x = y + 6$:
$$
x = \frac{52}{23} + 6 = \frac{52}{23} + \frac{138}{23} = \frac{190}{23}
$$

✔ Solution: $x = \frac{190}{23},\ y = \frac{52}{23}$

---

Problem 2:

$$
\begin{aligned}
(1)\quad & \frac{1}{3}x + \frac{1}{4}y = 5 \\
(2)\quad & \frac{1}{12}x + \frac{1}{3}y = 7
\end{aligned}
$$

To eliminate fractions, multiply both equations by the LCM of denominators (LCM of 3,4,12 is 12).

Multiply (1) by 12:
$$
12 \cdot \left(\frac{1}{3}x + \frac{1}{4}y\right) = 12 \cdot 5 \Rightarrow 4x + 3y = 60 \tag{1a}
$$

Multiply (2) by 12:
$$
12 \cdot \left(\frac{1}{12}x + \frac{1}{3}y\right) = 12 \cdot 7 \Rightarrow x + 4y = 84 \tag{2a}
$$

Now solve:
From (2a): $x = 84 - 4y$

Substitute into (1a):
$$
4(84 - 4y) + 3y = 60 \Rightarrow 336 - 16y + 3y = 60 \Rightarrow 336 - 13y = 60
$$

$$
-13y = 60 - 336 = -276 \Rightarrow y = \frac{276}{13}
$$

Now find $x$:
$$
x = 84 - 4 \cdot \frac{276}{13} = 84 - \frac{1104}{13} = \frac{1092}{13} - \frac{1104}{13} = -\frac{12}{13}
$$

✔ Solution: $x = -\frac{12}{13},\ y = \frac{276}{13}$

---

Problem 3:

$$
\begin{aligned}
(1)\quad & \frac{2}{3}x + \frac{5}{2}y = 100 \\
(2)\quad & \frac{1}{3}x + \frac{1}{2}y = 12
\end{aligned}
$$

Multiply both equations by 6 (LCM of 3 and 2):

Equation (1): $6 \cdot (\frac{2}{3}x + \frac{5}{2}y) = 6 \cdot 100 \Rightarrow 4x + 15y = 600$
Equation (2): $6 \cdot (\frac{1}{3}x + \frac{1}{2}y) = 6 \cdot 12 \Rightarrow 2x + 3y = 72$

Now we have:
$$
\begin{aligned}
(1a)\quad & 4x + 15y = 600 \\
(2a)\quad & 2x + 3y = 72
\end{aligned}
$$

Multiply (2a) by 2: $4x + 6y = 144$

Now subtract from (1a):
$$
(4x + 15y) - (4x + 6y) = 600 - 144 \Rightarrow 9y = 456 \Rightarrow y = \frac{456}{9} = \frac{152}{3}
$$

Now plug into (2a): $2x + 3(\frac{152}{3}) = 72 \Rightarrow 2x + 152 = 72 \Rightarrow 2x = -80 \Rightarrow x = -40$

✔ Solution: $x = -40,\ y = \frac{152}{3}$

---

Problem 4:

$$
\begin{aligned}
(1)\quad & \frac{3}{2}x + \frac{5}{3}y = 27 \\
(2)\quad & \frac{2}{3}x + \frac{5}{2}y = 7
\end{aligned}
$$

Multiply both equations to eliminate denominators.

LCM of 2 and 3 is 6.

Multiply (1) by 6:
$$
6 \cdot \left(\frac{3}{2}x + \frac{5}{3}y\right) = 6 \cdot 27 \Rightarrow 9x + 10y = 162 \tag{1a}
$$

Multiply (2) by 6:
$$
6 \cdot \left(\frac{2}{3}x + \frac{5}{2}y\right) = 6 \cdot 7 \Rightarrow 4x + 15y = 42 \tag{2a}
$$

Now solve:
We can use elimination. Let's make coefficients of $x$ match.

Multiply (1a) by 4: $36x + 40y = 648$
Multiply (2a) by 9: $36x + 135y = 378$

Now subtract:
$$
(36x + 135y) - (36x + 40y) = 378 - 648 \Rightarrow 95y = -270 \Rightarrow y = -\frac{270}{95} = -\frac{54}{19}
$$

Now plug into (2a): $4x + 15(-\frac{54}{19}) = 42$

$$
4x - \frac{810}{19} = 42 \Rightarrow 4x = 42 + \frac{810}{19} = \frac{798}{19} + \frac{810}{19} = \frac{1608}{19}
$$

$$
x = \frac{1608}{19 \cdot 4} = \frac{1608}{76} = \frac{402}{19}
$$

✔ Solution: $x = \frac{402}{19},\ y = -\frac{54}{19}$

---

Problem 5:

$$
\begin{aligned}
(1)\quad & \frac{2}{5}x + \frac{5}{3}y = 25 \\
(2)\quad & x - y = 10
\end{aligned}
$$

Solve (2): $x = y + 10$

Substitute into (1):
$$
\frac{2}{5}(y + 10) + \frac{5}{3}y = 25
\Rightarrow \frac{2}{5}y + 4 + \frac{5}{3}y = 25
$$

Common denominator for $y$: 15

$$
\frac{6}{15}y + \frac{25}{15}y = \frac{31}{15}y
\Rightarrow \frac{31}{15}y + 4 = 25
\Rightarrow \frac{31}{15}y = 21
\Rightarrow y = 21 \cdot \frac{15}{31} = \frac{315}{31}
$$

Then $x = y + 10 = \frac{315}{31} + \frac{310}{31} = \frac{625}{31}$

✔ Solution: $x = \frac{625}{31},\ y = \frac{315}{31}$

---

Problem 6:

$$
\begin{aligned}
(1)\quad & \frac{3}{5}x + \frac{2}{3}y = 27 \\
(2)\quad & \frac{2}{3}x + \frac{5}{7}y = 7
\end{aligned}
$$

This one has more complex denominators. Multiply both equations by LCM of denominators.

LCM of 5,3,7 is 105.

Multiply (1) by 105:
$$
105 \cdot \left(\frac{3}{5}x + \frac{2}{3}y\right) = 105 \cdot 27 \Rightarrow 63x + 70y = 2835 \tag{1a}
$$

Multiply (2) by 105:
$$
105 \cdot \left(\frac{2}{3}x + \frac{5}{7}y\right) = 105 \cdot 7 \Rightarrow 70x + 75y = 735 \tag{2a}
$$

Now solve:
Use elimination.

Make $x$ coefficients same. LCM of 63 and 70 is 900? Too big. Try smaller approach.

Let’s eliminate $y$. Coefficients: 70 and 75 → LCM is 2100.

Multiply (1a) by 30: $1890x + 2100y = 85050$
Multiply (2a) by 28: $1960x + 2100y = 20580$

Now subtract:
$$
(1960x + 2100y) - (1890x + 2100y) = 20580 - 85050
\Rightarrow 70x = -64470 \Rightarrow x = -\frac{64470}{70} = -921
$$

Wait — that seems too large. Let's double-check calculations.

Wait! Mistake in multiplication.

Let’s recompute:

(1a): $63x + 70y = 2835$
(2a): $70x + 75y = 735$

Let’s use substitution or better elimination.

Let’s eliminate $y$. Multiply (1a) by 15, (2a) by 14:

(1a) × 15: $945x + 1050y = 42525$
(2a) × 14: $980x + 1050y = 10290$

Now subtract:
$$
(980x + 1050y) - (945x + 1050y) = 10290 - 42525
\Rightarrow 35x = -32235 \Rightarrow x = -\frac{32235}{35} = -921
$$

Again, $x = -921$?

Plug into (2a): $70(-921) + 75y = 735$

$$
-64470 + 75y = 735 \Rightarrow 75y = 735 + 64470 = 65205
\Rightarrow y = \frac{65205}{75} = 869.4
$$

That’s not nice. Let's check original problem again.

Wait — maybe I made an error in multiplying.

Check (1): $\frac{3}{5}x + \frac{2}{3}y = 27$

Multiply by 105: $105 \cdot \frac{3}{5}x = 21 \cdot 3x = 63x$, correct.
$105 \cdot \frac{2}{3}y = 35 \cdot 2y = 70y$, correct.
$105 \cdot 27 = 2835$, correct.

(2): $\frac{2}{3}x + \frac{5}{7}y = 7$

$105 \cdot \frac{2}{3}x = 35 \cdot 2x = 70x$, correct.
$105 \cdot \frac{5}{7}y = 15 \cdot 5y = 75y$, correct.
$105 \cdot 7 = 735$, correct.

So equations are:
$$
63x + 70y = 2835 \quad \text{(1a)}\\
70x + 75y = 735 \quad \text{(2a)}
$$

Try solving via substitution.

From (1a): $63x + 70y = 2835$

Solve for $x$:
$$
63x = 2835 - 70y \Rightarrow x = \frac{2835 - 70y}{63}
$$

Plug into (2a):
$$
70 \cdot \left(\frac{2835 - 70y}{63}\right) + 75y = 735
$$

Multiply:
$$
\frac{70}{63}(2835 - 70y) + 75y = 735
\Rightarrow \frac{10}{9}(2835 - 70y) + 75y = 735
$$

Compute:
$$
\frac{10}{9} \cdot 2835 = 3150,\quad \frac{10}{9} \cdot (-70y) = -\frac{700}{9}y
$$

So:
$$
3150 - \frac{700}{9}y + 75y = 735
\Rightarrow 3150 + \left(75 - \frac{700}{9}\right)y = 735
$$

Convert 75 to ninths: $\frac{675}{9}$

$$
\left(\frac{675 - 700}{9}\right)y = -\frac{25}{9}y
\Rightarrow 3150 - \frac{25}{9}y = 735
\Rightarrow -\frac{25}{9}y = 735 - 3150 = -2415
\Rightarrow \frac{25}{9}y = 2415
\Rightarrow y = 2415 \cdot \frac{9}{25} = \frac{21735}{25} = 869.4
$$

Same as before.

But $y = 869.4 = \frac{4347}{5}$

Then $x = \frac{2835 - 70 \cdot \frac{4347}{5}}{63} = \frac{2835 - 60858}{63} = \frac{-58023}{63} = -921$

So $x = -921,\ y = 869.4$

But this seems very large. Let’s verify if it satisfies original equation.

Check (1): $\frac{3}{5}(-921) + \frac{2}{3}(869.4) = ?$

$\frac{3}{5} \cdot -921 = -552.6$

$\frac{2}{3} \cdot 869.4 = 579.6$

Sum: $-552.6 + 579.6 = 27$ ✔

Check (2): $\frac{2}{3}(-921) + \frac{5}{7}(869.4)$

$\frac{2}{3} \cdot -921 = -614$

$\frac{5}{7} \cdot 869.4 ≈ 621$

$-614 + 621 = 7$ ✔

So it’s correct, even though numbers are large.

✔ Solution: $x = -921,\ y = \frac{4347}{5}$ or $869.4$

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✔ Final Answers:

1. $x = \frac{190}{23},\ y = \frac{52}{23}$
2. $x = -\frac{12}{13},\ y = \frac{276}{13}$
3. $x = -40,\ y = \frac{152}{3}$
4. $x = \frac{402}{19},\ y = -\frac{54}{19}$
5. $x = \frac{625}{31},\ y = \frac{315}{31}$
6. $x = -921,\ y = \frac{4347}{5}$

Let me know if you want decimal approximations or simplified forms!