Question

1. does the point (- 5, - 20) lie on the line $y = 6x + 10$? explain
2. does the point (6, - 10) lie on the line $y = -\frac{5}{3}x - 4$? explain
3. the point (2, a) is on the line $y = 5x - 30$. determine the value of a.
4. identify whether each of the following linear equations is in standard form. if not, fix it.

a) $2x - 3y - 6 = 0$
e) $\frac{1}{2}x + y - 5 = 0$
b) $5x - 2y = - 3$
f) $x + 2y - 10 = 0$
c) $y = 2x + 12$
g) $- 4x - 9y + 4 = 0$
d) $5y + 2x + 25 = 0$

1. express each of the following equations in slope y-intercept form.

a) $7x + y - 4 = 0$
b) $3x + 2y - 8 = 0$
c) $x - 4y - 2 = 0$

Explanation:

Question 4

Step1: Substitute x and y values

Substitute \( x = -5 \) and \( y = -20 \) into the equation \( y = 6x + 10 \).
Left side (y): \( -20 \)
Right side: \( 6(-5) + 10 = -30 + 10 = -20 \)

Step2: Compare both sides

Since left side (\( -20 \)) equals right side (\( -20 \)), the point lies on the line.

Step1: Substitute x and y values

Substitute \( x = 6 \) and \( y = -10 \) into the equation \( y = -\frac{5}{3}x - 4 \).
Left side (y): \( -10 \)
Right side: \( -\frac{5}{3}(6) - 4 = -10 - 4 = -14 \)

Step2: Compare both sides

Since \( -10
eq -14 \), the point does not lie on the line.

Step1: Substitute x value

The point \((2, a)\) is on the line \( y = 5x - 30 \), so substitute \( x = 2 \) into the equation.
\( a = 5(2) - 30 \)

Step2: Calculate a

\( a = 10 - 30 = -20 \)

Answer:

Yes, the point \((-5, -20)\) lies on the line \( y = 6x + 10 \) because substituting \( x = -5 \) and \( y = -20 \) into the equation gives \( -20 = 6(-5) + 10 \), which simplifies to \( -20 = -20 \), a true statement.

Question 5