We are asked to solve two systems of linear equations without using graphing. System a: $2x + 3y = 5$ $2x + 4y = 9$ System b: $\frac{2}{3}x + y = \frac{7}{3}$ $\frac{2}{3}x - y = 1$

1. Problem Description

We are asked to solve two systems of linear equations without using graphing.

System a:

$2x + 3y = 5$

$2x + 4y = 9$

System b:

$\frac{2}{3}x + y = \frac{7}{3}$

$\frac{2}{3}x - y = 1$

2. Solution Steps

a. Solve the first system:

$2x + 3y = 5$

$2x + 4y = 9$

We can use the elimination method. Subtract the first equation from the second equation:

$(2x + 4y) - (2x + 3y) = 9 - 5$

$2x + 4y - 2x - 3y = 4$

$y = 4$

Substitute $y = 4$ into the first equation:

$2x + 3(4) = 5$

$2x + 12 = 5$

$2x = 5 - 12$

$2x = -7$

$x = -\frac{7}{2}$

So the solution for system a is $x = -\frac{7}{2}$ and $y = 4$.

b. Solve the second system:

$\frac{2}{3}x + y = \frac{7}{3}$

$\frac{2}{3}x - y = 1$

We can use the elimination method. Add the two equations:

$(\frac{2}{3}x + y) + (\frac{2}{3}x - y) = \frac{7}{3} + 1$

$\frac{2}{3}x + y + \frac{2}{3}x - y = \frac{7}{3} + \frac{3}{3}$

$\frac{4}{3}x = \frac{10}{3}$

$x = \frac{10}{3} \cdot \frac{3}{4}$

$x = \frac{10}{4}$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ into the first equation:

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

$\frac{5}{3} + y = \frac{7}{3}$

$y = \frac{7}{3} - \frac{5}{3}$

$y = \frac{2}{3}$

So the solution for system b is $x = \frac{5}{2}$ and $y = \frac{2}{3}$.

3. Final Answer

a. $x = -\frac{7}{2}, y = 4$

b. $x = \frac{5}{2}, y = \frac{2}{3}$