Question

1. $y - x = 4$ 11. $x = \frac{2}{5}y$ 12. $y + 3 = x + 6$ 13. $y - 5 = 2x$ 14. $x - y = 0$ 15. $\frac{x}{y} = 2$ 16. $8 = xy$ 17. $x^2 = y$

Explanation:

Step1: Identify the problem type

These are equations involving two variables \(x\) and \(y\). The likely task is to analyze them (e.g., determine if they are linear, solve for one variable in terms of the other, etc.). Let's take equation 10: \(y - x=4\).

Step2: Solve for \(y\) in terms of \(x\)

Add \(x\) to both sides: \(y=x + 4\).

Step3: Equation 11: \(x=\frac{2}{5}y\)

Solve for \(y\): Multiply both sides by \(\frac{5}{2}\), so \(y=\frac{5}{2}x\).

Step4: Equation 12: \(y + 3=x + 6\)

Subtract 3 from both sides: \(y=x+6 - 3=x + 3\).

Step5: Equation 13: \(y - 5=2x\)

Add 5 to both sides: \(y=2x + 5\).

Step6: Equation 14: \(x - y=0\)

Add \(y\) to both sides: \(x=y\) or \(y=x\).

Step7: Equation 15: \(\frac{x}{y}=2\)

Multiply both sides by \(y\) (assuming \(y
eq0\)): \(x = 2y\), then solve for \(y\): \(y=\frac{1}{2}x\).

Step8: Equation 16: \(8=xy\)

Solve for \(y\) (assuming \(x
eq0\)): \(y=\frac{8}{x}\).

Step9: Equation 17: \(x^{2}=y\)

This is a quadratic relationship, \(y=x^{2}\).

Answer:

For each equation, solving for \(y\) in terms of \(x\) (or vice - versa) gives:

1. \(y=x + 4\)
2. \(y=\frac{5}{2}x\)
3. \(y=x + 3\)
4. \(y=2x + 5\)
5. \(y=x\)
6. \(y=\frac{1}{2}x\) (for \(y

eq0\))

1. \(y=\frac{8}{x}\) (for \(x

eq0\))

1. \(y=x^{2}\)