Question

1. algebra in the figure at the right, m∠pqr = 4x + 47. find m∠pqs. 22. algebra points a, b, and c are collinear with b between a and c. ab = 4x - 1, bc = 2x + 1, and ac = 8x - 4. find ab, bc, and ac.

Explanation:

21.

Step1: Note angle - relationship

Since $\angle PQR=\angle PQS + \angle SQR$, and $\angle SQR = 36^{\circ}$, $\angle PQR=4x + 47$ and $\angle PQS=7x - 4$.
So, $4x + 47=(7x - 4)+36$.

Step2: Solve the equation for $x$

First, simplify the right - hand side of the equation: $(7x - 4)+36=7x+32$.
The equation becomes $4x + 47=7x+32$.
Subtract $4x$ from both sides: $47 = 3x+32$.
Subtract 32 from both sides: $3x=47 - 32=15$.
Divide both sides by 3: $x = 5$.

Step3: Find $m\angle PQS$

Substitute $x = 5$ into the expression for $\angle PQS$: $m\angle PQS=7x - 4$.
$m\angle PQS=7\times5-4=35 - 4=31^{\circ}$.

Step1: Use the collinear - point property

Since $A$, $B$, and $C$ are collinear with $B$ between $A$ and $C$, then $AB + BC=AC$.
Substitute $AB = 4x - 1$, $BC = 2x + 1$, and $AC = 8x - 4$ into the equation: $(4x - 1)+(2x + 1)=8x - 4$.

Step2: Simplify the left - hand side of the equation

$(4x - 1)+(2x + 1)=4x+2x-1 + 1=6x$.
The equation becomes $6x=8x - 4$.

Step3: Solve the equation for $x$

Subtract $6x$ from both sides: $0=8x-6x - 4$, so $2x-4 = 0$.
Add 4 to both sides: $2x=4$.
Divide both sides by 2: $x = 2$.

Step4: Find $AB$, $BC$, and $AC$

For $AB$: Substitute $x = 2$ into $AB = 4x - 1$, $AB=4\times2-1=7$.
For $BC$: Substitute $x = 2$ into $BC = 2x + 1$, $BC=2\times2 + 1=5$.
For $AC$: Substitute $x = 2$ into $AC = 8x - 4$, $AC=8\times2-4=12$.

Answer:

$m\angle PQS = 31^{\circ}$

22.