3. Solve for $$ x $$ and $$ y $$ using any method of your choice: (a) $$ x+y=1 $$ and $$ x-2 y=1 $$ (b) $$ 3 x+2 y=2 $$ and $$ 5 x-2 y=-18 $$ (c) $$ x+4 y=14 $$ and $$ 3 x+2 y=12 $$ (d) $$ 2 y-3 x=7 $$ and $$ 4 y-5 x=21 $$ (e) $$ 3 x+2 y=6 $$ and $$ 5 x+3 y=11 $$ (d) $$ \frac{y-2}{3}-\frac{x-1}{2}=1 $$ and $$ \frac{y-4}{5}-\frac{x-3}{4}=1 $$

Question

UpStudy AI · Accepted Answer

### (a) Solve the system  
$$
x+y=1,\quad x-2y=1.
$$

1. Express $$ x $$ from the first equation:  
   $$
   x=1-y.
   $$
2. Substitute in the second equation:  
   $$
   (1-y)-2y=1 \quad\Longrightarrow\quad 1-3y=1.
   $$
3. Solve for $$ y $$:  
   $$
   -3y=0 \quad\Longrightarrow\quad y=0.
   $$
4. Then $$ x=1-0=1 $$.

$$
\boxed{x=1,\quad y=0}
$$

---

### (b) Solve the system  
$$
3x+2y=2,\quad 5x-2y=-18.
$$

1. Add the equations to eliminate $$ y $$:  
   $$
   (3x+2y)+(5x-2y)=2+(-18) \quad\Longrightarrow\quad 8x=-16.
   $$
2. Solve for $$ x $$:  
   $$
   x=-2.
   $$
3. Substitute $$ x=-2 $$ into $$ 3x+2y=2 $$:  
   $$
   3(-2)+2y=2 \quad\Longrightarrow\quad -6+2y=2.
   $$
4. Solve for $$ y $$:  
   $$
   2y=8 \quad\Longrightarrow\quad y=4.
   $$

$$
\boxed{x=-2,\quad y=4}
$$

---

### (c) Solve the system  
$$
x+4y=14,\quad 3x+2y=12.
$$

1. Express $$ x $$ from the first equation:  
   $$
   x=14-4y.
   $$
2. Substitute into the second equation:  
   $$
   3(14-4y)+2y=12 \quad\Longrightarrow\quad 42-12y+2y=12.
   $$
3. Simplify and solve for $$ y $$:  
   $$
   42-10y=12 \quad\Longrightarrow\quad -10y=12-42=-30, \quad y=3.
   $$
4. Then find $$ x $$:  
   $$
   x=14-4(3)=14-12=2.
   $$

$$
\boxed{x=2,\quad y=3}
$$

---

### (d) Solve the system  
$$
2y-3x=7,\quad 4y-5x=21.
$$

1. Multiply the first equation by $$ 2 $$ to align the $$ y $$-terms:  
   $$
   4y-6x=14.
   $$
2. Subtract the second equation from this result:  
   $$
   (4y-6x)-(4y-5x)=14-21 \quad\Longrightarrow\quad -x=-7.
   $$
3. Solve for $$ x $$:  
   $$
   x=7.
   $$
4. Substitute $$ x=7 $$ into the first equation:  
   $$
   2y-3(7)=7 \quad\Longrightarrow\quad 2y-21=7.
   $$
5. Solve for $$ y $$:  
   $$
   2y=28 \quad\Longrightarrow\quad y=14.
   $$

$$
\boxed{x=7,\quad y=14}
$$

---

### (e) Solve the system  
$$
3x+2y=6,\quad 5x+3y=11.
$$

1. Multiply the first equation by $$ 3 $$ and the second by $$ 2 $$ to eliminate $$ y $$:  
   $$
   9x+6y=18,\quad 10x+6y=22.
   $$
2. Subtract the first from the second:  
   $$
   (10x+6y)-(9x+6y)=22-18 \quad\Longrightarrow\quad x=4.
   $$
3. Substitute $$ x=4 $$ into $$ 3x+2y=6 $$:  
   $$
   3(4)+2y=6 \quad\Longrightarrow\quad 12+2y=6.
   $$
4. Solve for $$ y $$:  
   $$
   2y=6-12=-6\quad\Longrightarrow\quad y=-3.
   $$

$$
\boxed{x=4,\quad y=-3}
$$

---

### (f) Solve the system  
$$
\frac{y-2}{3}-\frac{x-1}{2}=1,\quad \frac{y-4}{5}-\frac{x-3}{4}=1.
$$

**Step 1. Multiply to Eliminate Denominators in the First Equation:**  
The LCM of $$3$$ and $$2$$ is $$6$$. Multiply the first equation by $$6$$:  
$$
6\left(\frac{y-2}{3}\right)-6\left(\frac{x-1}{2}\right)=6 \quad\Longrightarrow\quad 2(y-2)-3(x-1)=6.
$$
Expand:  
$$
2y-4-3x+3=6 \quad\Longrightarrow\quad 2y-3x-1=6.
$$
Add $$1$$ to both sides:  
$$
2y-3x=7.
$$

**Step 2. Multiply to Eliminate Denominators in the Second Equation:**  
The LCM of $$5$$ and $$4$$ is $$20$$. Multiply the second equation by $$20$$:  
$$
20\left(\frac{y-4}{5}\right)-20\left(\frac{x-3}{4}\right)=20 \quad\Longrightarrow\quad 4(y-4)-5(x-3)=20.
$$
Expand:  
$$
4y-16-5x+15=20 \quad\Longrightarrow\quad 4y-5x-1=20.
$$
Add $$1$$ to both sides:  
$$
4y-5x=21.
$$

The resulting system is identical to the one in part (d):  
$$
2y-3x=7,\quad 4y-5x=21.
$$
We already solved this system in part (d) with  
$$
\boxed{x=7,\quad y=14}.
$$

---

### Final Answers

(a) $$\boxed{x=1,\; y=0}$$  
(b) $$\boxed{x=-2,\; y=4}$$  
(c) $$\boxed{x=2,\; y=3}$$  
(d) $$\boxed{x=7,\; y=14}$$  
(e) $$\boxed{x=4,\; y=-3}$$  
(f) $$\boxed{x=7,\; y=14}$$
$$ \begin{align*} (a) & \quad x=1,\; y=0 \ (b) & \quad x=-2,\; y=4 \ (c) & \quad x=2,\; y=3 \ (d) & \quad x=7,\; y=14 \ (e) & \quad x=4,\; y=-3 \ (f) & \quad x=7,\; y=14 \ \end{align*} $$

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