Question

unit 4: linear relations

1. which graph represents the equation $y = 2x + 3$?

graph with x-axis from 0 to 9, y-axis from 0 to 11, and lines p, q, r, s
a. line s b. line q c. line p d. line r
unit 5: polynomials

1. subtract: $(3x - 7x^2 + 2) - (4x^2 - 5 + 6x)$

a. $-11x^2 + 3x - 7$ c. $-11x^2 - 3x + 7$
b. $-11x^2 - 9x - 3$ d. $11x^2 + 3x - 7$

1. subtract: $(3y^2 - 5x^2 + 4) - (2x - 8 + 4y^2)$

a. $-1y^2 - 5x^2 - 2x - 4$ c. $-4x + 12$
b. $3y^2 - 7x^2 + 12$ d. $-1y^2 - 5x^2 - 2x + 12$
questions i should practice:
/3

Explanation:

Question 11

Step1: Recall slope-intercept form

The equation \( y = 2x + 3 \) is in slope - intercept form \( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept. Here, \( m = 2 \) (slope) and \( b=3 \) (y - intercept, the value of \( y \) when \( x = 0 \)).

Step2: Analyze the y - intercept

For the line with equation \( y = 2x+3 \), when \( x = 0 \), \( y=3 \). So the line should cross the y - axis at \( (0,3) \).

Step3: Analyze the slope

The slope \( m = 2=\frac{\text{rise}}{\text{run}}=\frac{2}{1} \). This means for every 1 unit we move to the right (increase in \( x \) by 1), we move up 2 units (increase in \( y \) by 2).
Looking at the lines:

• Line P: Let's check its y - intercept and slope. If we assume the y - intercept is around 3 and the slope is 2 (since from \( x = 0,y = 3 \), when \( x = 1,y=5 \) (increase of 2 in \( y \) for increase of 1 in \( x \))), this matches \( y = 2x + 3 \).
• Line Q: Its y - intercept and slope do not match \( y=2x + 3 \).
• Line S: Its y - intercept is 0, not 3.
• Line R: Its y - intercept and slope do not match \( y = 2x+3 \).

Step1: Distribute the negative sign

We have \( (3x-7x^{2}+2)-(4x^{2}-5 + 6x)=3x-7x^{2}+2-4x^{2}+5 - 6x \) (because when we subtract a polynomial, we distribute the negative sign to each term inside the parentheses: \( -(4x^{2}-5 + 6x)=-4x^{2}+5 - 6x \))

Step2: Combine like terms

• For the \( x^{2} \) terms: \( -7x^{2}-4x^{2}=-11x^{2} \)
• For the \( x \) terms: \( 3x-6x=-3x \)
• For the constant terms: \( 2 + 5=7 \)

Putting it all together: \( -11x^{2}-3x + 7 \)

Step1: Distribute the negative sign

\( (3y^{2}-5x^{2}+4)-(2x - 8+4y^{2})=3y^{2}-5x^{2}+4-2x + 8-4y^{2} \) (distribute the negative sign to each term in \( (2x - 8+4y^{2}) \), so \( -(2x - 8+4y^{2})=-2x + 8-4y^{2} \))

Step2: Combine like terms

• For the \( y^{2} \) terms: \( 3y^{2}-4y^{2}=-y^{2} \) (or \( -1y^{2} \))
• For the \( x^{2} \) terms: \( -5x^{2} \) (no other \( x^{2} \) terms)
• For the \( x \) terms: \( -2x \) (no other \( x \) terms)
• For the constant terms: \( 4 + 8 = 12 \)

Combining these, we get \( -y^{2}-5x^{2}-2x + 12 \) (or \( -1y^{2}-5x^{2}-2x + 12 \))

Answer:

c. Line P

Question 12