Solve the following systems of equations using the elimination method: 1. 1/2x + 1/3y = 4, 1/2x − 1/3y = 2 2…

Question

Solve the following systems of equations using the elimination method: 1. 1/2x + 1/3y = 4, 1/2x − 1/3y = 2 2. 2/3x + 1/4y = 5, 2/3x − 1/4y = 1 3. 1/2x + 1/4y

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the systems of equations using the elimination method.

1.
Given the system:

\frac{1}{2}x + \frac{1}{3}y = 4 \quad (1)

\frac{1}{2}x - \frac{1}{3}y = 2 \quad (2)

Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{1}{2}x + \frac{1}{3}y\right) + \left(\frac{1}{2}x - \frac{1}{3}y\right) = 4 + 2$ $\frac{1}{2}x + \frac{1}{2}x = 6$ $x = 6$
Substitute $x=6$ into equation (1).
$\frac{1}{2}(6) + \frac{1}{3}y = 4$ $3 + \frac{1}{3}y = 4$ $\frac{1}{3}y = 4 - 3$ $\frac{1}{3}y = 1$ $y = 3$
The solution is $(x, y) = (6, 3)$ .
$\boxed{x=6, y=3}$
2.
Given the system:
$\frac{2}{3}x + \frac{1}{4}y = 5 \quad (1)$ $\frac{2}{3}x - \frac{1}{4}y = 1 \quad (2)$
Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{2}{3}x + \frac{1}{4}y\right) + \left(\frac{2}{3}x - \frac{1}{4}y\right) = 5 + 1$ $\frac{2}{3}x + \frac{2}{3}x = 6$ $\frac{4}{3}x = 6$ $x = 6 \times \frac{3}{4}$ $x = \frac{18}{4} = \frac{9}{2}$
Substitute $x=\frac{9}{2}$ into equation (1).
$\frac{2}{3}\left(\frac{9}{2}\right) + \frac{1}{4}y = 5$ $3 + \frac{1}{4}y = 5$ $\frac{1}{4}y = 5 - 3$ $\frac{1}{4}y = 2$ $y = 8$
The solution is $(x, y) = \left(\frac{9}{2}, 8\right)$ .
$\boxed{x=\frac{9}{2}, y=8}$
3.
Given the system:
$\frac{1}{2}x + \frac{1}{4}y = 3 \quad (1)$ $\frac{1}{2}x - \frac{1}{5}y = 1 \quad (2)$
Subtract equation (2) from equation (1) to eliminate $x$ .
$\left(\frac{1}{2}x + \frac{1}{4}y\right) - \left(\frac{1}{2}x - \frac{1}{5}y\right) = 3 - 1$ $\frac{1}{4}y + \frac{1}{5}y = 2$ $\left(\frac{5}{20} + \frac{4}{20}\right)y = 2$ $\frac{9}{20}y = 2$ $y = 2 \times \frac{20}{9}$ $y = \frac{40}{9}$
Substitute $y=\frac{40}{9}$ into equation (1).
$\frac{1}{2}x + \frac{1}{4}\left(\frac{40}{9}\right) = 3$ $\frac{1}{2}x + \frac{10}{9} = 3$ $\frac{1}{2}x = 3 - \frac{10}{9}$ $\frac{1}{2}x = \frac{27}{9} - \frac{10}{9}$ $\frac{1}{2}x = \frac{17}{9}$ $x = 2 \times \frac{17}{9}$ $x = \frac{34}{9}$
The solution is $(x, y) = \left(\frac{34}{9}, \frac{40}{9}\right)$ .
$\boxed{x=\frac{34}{9}, y=\frac{40}{9}}$
4.
Given the system:
$\frac{3}{4}x + \frac{1}{2}y = 6 \quad (1)$ $\frac{3}{4}x - \frac{1}{2}y = 2 \quad (2)$
Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{3}{4}x + \frac{1}{2}y\right) + \left(\frac{3}{4}x - \frac{1}{2}y\right) = 6 + 2$ $\frac{3}{4}x + \frac{3}{4}x = 8$ $\frac{6}{4}x = 8$ $\frac{3}{2}x = 8$ $x = 8 \times \frac{2}{3}$ $x = \frac{16}{3}$
Substitute $x=\frac{16}{3}$ into equation (1).
$\frac{3}{4}\left(\frac{16}{3}\right) + \frac{1}{2}y = 6$ $4 + \frac{1}{2}y = 6$ $\frac{1}{2}y = 6 - 4$ $\frac{1}{2}y = 2$ $y = 4$
The solution is $(x, y) = \left(\frac{16}{3}, 4\right)$ .
$\boxed{x=\frac{16}{3}, y=4}$
5.
Given the system:
$\frac{2}{5}x + \frac{3}{5}y = 7 \quad (1)$ $\frac{2}{5}x - \frac{3}{5}y = 1 \quad (2)$
Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{2}{5}x + \frac{3}{5}y\right) + \left(\frac{2}{5}x - \frac{3}{5}y\right) = 7 + 1$ $\frac{2}{5}x + \frac{2}{5}x = 8$ $\frac{4}{5}x = 8$ $x = 8 \times \frac{5}{4}$ $x = 10$
Substitute $x=10$ into equation (1).
$\frac{2}{5}(10) + \frac{3}{5}y = 7$ $4 + \frac{3}{5}y = 7$ $\frac{3}{5}y = 7 - 4$ $\frac{3}{5}y = 3$ $y = 3 \times \frac{5}{3}$ $y = 5$
The solution is $(x, y) = (10, 5)$ .
$\boxed{x=10, y=5}$
6.
Given the system:
$\frac{1}{3}x + \frac{1}{2}y = 5 \quad (1)$ $\frac{1}{6}x + \frac{1}{2}y = 4 \quad (2)$
Subtract equation (2) from equation (1) to eliminate $y$ .
$\left(\frac{1}{3}x + \frac{1}{2}y\right) - \left(\frac{1}{6}x + \frac{1}{2}y\right) = 5 - 4$ $\frac{1}{3}x - \frac{1}{6}x = 1$ $\left(\frac{2}{6} - \frac{1}{6}\right)x = 1$ $\frac{1}{6}x = 1$ $x = 6$
Substitute $x=6$ into equation (1).
$\frac{1}{3}(6) + \frac{1}{2}y = 5$ $2 + \frac{1}{2}y = 5$ $\frac{1}{2}y = 5 - 2$ $\frac{1}{2}y = 3$ $y = 6$
The solution is $(x, y) = (6, 6)$ .
$\boxed{x=6, y=6}$
7.
Given the system:
$\frac{3}{5}x + \frac{1}{3}y = 8 \quad (1)$ $\frac{3}{5}x - \frac{1}{3}y = 2 \quad (2)$
Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{3}{5}x + \frac{1}{3}y\right) + \left(\frac{3}{5}x - \frac{1}{3}y\right) = 8 + 2$ $\frac{3}{5}x + \frac{3}{5}x = 10$ $\frac{6}{5}x = 10$ $x = 10 \times \frac{5}{6}$ $x = \frac{50}{6} = \frac{25}{3}$
Substitute $x=\frac{25}{3}$ into equation (1).
$\frac{3}{5}\left(\frac{25}{3}\right) + \frac{1}{3}y = 8$ $5 + \frac{1}{3}y = 8$ $\frac{1}{3}y = 8 - 5$ $\frac{1}{3}y = 3$ $y = 9$
The solution is $(x, y) = \left(\frac{25}{3}, 9\right)$ .
$\boxed{x=\frac{25}{3}, y=9}$
8.
Given the system:
$\frac{1}{4}x + \frac{1}{3}y = 3 \quad (1)$ $\frac{1}{4}x - \frac{1}{3}y = 1 \quad (2)$
Add equation (1) and equation (2) to eliminate $y$ .
$\left(\frac{1}{4}x + \frac{1}{3}y\right) + \left(\frac{1}{4}x - \frac{1}{3}y\right) = 3 + 1$ $\frac{1}{4}x + \frac{1}{4}x = 4$ $\frac{2}{4}x = 4$ $\frac{1}{2}x = 4$ $x = 8$
Substitute $x=8$ into equation (1).
$\frac{1}{4}(8) + \frac{1}{3}y = 3$ $2 + \frac{1}{3}y = 3$ $\frac{1}{3}y = 3 - 2$ $\frac{1}{3}y = 1$ $y = 3$
The solution is $(x, y) = (8, 3)$ .
$\boxed{x=8, y=3}$

Solve the following systems of equations using the elimination method: 1. 1/2x + 1/3y = 4, 1/2x - 1/3y = 2 2. 2/3x + 1/4y = 5, 2/3x - 1/4y = 1 3. 1/2x + 1/4y

Step-by-step solution

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=6$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{9}{2}$ into equation (1).

Subtract equation (2) from equation (1) to eliminate $x$ .

Substitute $y=\frac{40}{9}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{16}{3}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=10$ into equation (1).

Subtract equation (2) from equation (1) to eliminate $y$ .

Substitute $x=6$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{25}{3}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=8$ into equation (1).

Still stuck on this one?

More Mathematics Questions

Solve the following systems of equations using the elimination method: 1. 1/2x + 1/3y = 4, 1/2x - 1/3y = 2 2. 2/3x + 1/4y = 5, 2/3x - 1/4y = 1 3. 1/2x + 1/4y

Step-by-step solution

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=6x=6x=6 into equation (1).

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=92x=\frac{9}{2}x=29​ into equation (1).

Subtract equation (2) from equation (1) to eliminate xxx.

Substitute y=409y=\frac{40}{9}y=940​ into equation (1).

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=163x=\frac{16}{3}x=316​ into equation (1).

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=10x=10x=10 into equation (1).

Subtract equation (2) from equation (1) to eliminate yyy.

Substitute x=6x=6x=6 into equation (1).

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=253x=\frac{25}{3}x=325​ into equation (1).

Add equation (1) and equation (2) to eliminate yyy.

Substitute x=8x=8x=8 into equation (1).

Still stuck on this one?

More Mathematics Questions

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=6$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{9}{2}$ into equation (1).

Subtract equation (2) from equation (1) to eliminate $x$ .

Substitute $y=\frac{40}{9}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{16}{3}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=10$ into equation (1).

Subtract equation (2) from equation (1) to eliminate $y$ .

Substitute $x=6$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=\frac{25}{3}$ into equation (1).

Add equation (1) and equation (2) to eliminate $y$ .

Substitute $x=8$ into equation (1).