solve using subsitution, elimination or the transitive property: y=-9/10x+3 and y=-4/3+3


You provided two equations:

1. \( y = -\frac{9}{10}x + 3 \)
2. \( y = -\frac{4}{3} + 3 \)

Let's solve this system using substitution, since both equations are already in terms of \( y \).

### Step 1: Set the equations equal to each other
Since both expressions equal \( y \), we can set the right-hand sides equal to each other:

\[
-\frac{9}{10}x + 3 = -\frac{4}{3} + 3
\]

### Step 2: Simplify both sides
On the right-hand side, simplify the constants:

\[
-\frac{4}{3} + 3 = -\frac{4}{3} + \frac{9}{3} = \frac{5}{3}
\]

Now the equation becomes:

\[
-\frac{9}{10}x + 3 = \frac{5}{3}
\]

### Step 3: Subtract 3 from both sides to isolate the term with \( x \):

\[
-\frac{9}{10}x = \frac{5}{3} - 3
\]

Simplifying the right-hand side:

\[
-\frac{9}{10}x = \frac{5}{3} - \frac{9}{3} = -\frac{4}{3}
\]

### Step 4: Solve for \( x \)
Now, solve for \( x \) by multiplying both sides by the reciprocal of \( -\frac{9}{10} \), which is \( -\frac{10}{9} \):

\[
x = \left( -\frac{4}{3} \right) \times \left( -\frac{10}{9} \right)
\]

\[
x = \frac{40}{27}
\]

### Step 5: Substitute \( x = \frac{40}{27} \) back into one of the original equations
Let's use the second equation:

\[
y = -\frac{4}{3} + 3
\]

We already simplified this equation earlier:

\[
y = \frac{5}{3}
\]

Thus, the solution to the system of equations is:

\[
x = \frac{40}{27}, \quad y = \frac{5}{3}
\]

Would you like any further details or have any questions?

Here are 5 related questions:
1. How do you check if the solution satisfies both equations?
2. What are the advantages of using substitution over elimination in this case?
3. How does solving the system graphically compare to algebraic methods?
4. What happens if the two lines were parallel? How would that affect the solution?
5. Can we use a matrix approach (Gaussian elimination) to solve this system?

**Tip:** Always simplify both sides of the equation as much as possible before attempting to isolate variables.

solve using substitution, elimination or the transitive property: y=-9/10x+3 and y=-4/3+3

Learn how to solve a system of linear equations using substitution, elimination, or the transitive property. This problem covers key algebraic concepts, including solving y = -9/10x + 3 and y = -4/3 + 3 step by step. Suitable for students in Grades 8-10.

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