solving/evaluating
part 3: algebra
1. if df = 42, find de.
2. in the diagram below, if jl = 10x - 2, jk = 5x - 8, and kl = 7x - 12, find kl.
3. if s is the mid - point of rt, rs = 5x + 17, and st = 8x - 31, find rs.
4. if line y bisects ac, ab = 4 - 5x, and bc = 2x + 25, find ac.
5. if b is the mid - point of ac, ac = cd, ab = 3x + 4, ac = 11x - 17, and ce = 49, find de.
6. if m∠def = 117°, find the value of x.
7. if m∠pqs = 16°, m∠sqr=(9x + 17)°, and m∠pqr=(12x - 6)°, find m∠pqr.
8. if ef bisects ∠aed, m∠aef=(4x + 3)°, and m∠fed=(7x - 33)°, find m∠ceb.
9. ∠1 and ∠2 form a linear pair. if m∠1=(18x - 1)° and m∠2=(23x + 17)°, find m∠2.
11. ∠1 and ∠2 are vertical angles. if m∠1=(5x + 12)° and m∠2=(6x - 11)°, find m∠1.

Question

Accepted Answer

### 1. If $DF = 42$, find $DE$.
# Explanation:
## Step1: Set up equation based on segment - addition
Since $DF=DE + EF$, and $DE = 7x + 1$, $EF=4x - 3$, we have $DF=(7x + 1)+(4x - 3)$. Also, $DF = 42$, so $(7x + 1)+(4x - 3)=42$.
$$7x+1 + 4x-3=42$$
$$11x-2 = 42$$
## Step2: Solve for $x$
Add 2 to both sides of the equation $11x-2 = 42$:
$$11x=42 + 2=44$$
Divide both sides by 11: $x = 4$.
## Step3: Find $DE$
Substitute $x = 4$ into the expression for $DE$: $DE=7x + 1$.
$$DE=7\times4+1=28 + 1=29$$

### 2. In the diagram below, if $JL=10x - 2$, $JK = 5x - 8$, and $KL=7x - 12$, find $KL$.
# Explanation:
## Step1: Set up equation based on segment - addition
Since $JL=JK + KL$, we have $10x-2=(5x - 8)+(7x - 12)$.
$$10x-2=5x-8 + 7x-12$$
$$10x-2=12x-20$$
## Step2: Solve for $x$
Subtract $10x$ from both sides: $-2=12x-10x-20$, so $-2 = 2x-20$.
Add 20 to both sides: $2x=-2 + 20 = 18$.
Divide both sides by 2: $x = 9$.
## Step3: Find $KL$
Substitute $x = 9$ into the expression for $KL$: $KL=7x - 12$.
$$KL=7\times9-12=63-12=51$$

### 3. If $S$ is the mid - point of $\overline{RT}$, $RS = 5x+17$, and $ST = 8x - 31$, find $RS$.
# Explanation:
## Step1: Set up equation based on mid - point property
Since $S$ is the mid - point of $\overline{RT}$, $RS = ST$. So $5x+17=8x - 31$.
## Step2: Solve for $x$
Subtract $5x$ from both sides: $17=8x-5x - 31$, so $17 = 3x-31$.
Add 31 to both sides: $3x=17 + 31=48$.
Divide both sides by 3: $x = 16$.
## Step3: Find $RS$
Substitute $x = 16$ into the expression for $RS$: $RS=5x+17$.
$$RS=5\times16+17=80 + 17=97$$

### 4. If line $y$ bisects $\overline{AC}$, $AB = 4-5x$, and $BC=2x + 25$, find $AC$.
# Explanation:
## Step1: Set up equation based on bisect property
Since line $y$ bisects $\overline{AC}$, $AB = BC$. So $4-5x=2x + 25$.
## Step2: Solve for $x$
Add $5x$ to both sides: $4=2x+5x + 25$, so $4 = 7x+25$.
Subtract 25 from both sides: $7x=4 - 25=-21$.
Divide both sides by 7: $x=-3$.
## Step3: Find $AC$
First, find $AB$ or $BC$. Let's find $AB$: $AB = 4-5\times(-3)=4 + 15 = 19$. Since $AC=2AB$ (because $AB = BC$), $AC=2\times19 = 38$.

### 5. If $B$ is the mid - point of $\overline{AC}$, $AC = CD$, $AB = 3x + 4$, $AC=11x - 17$, and $CE = 49$, find $DE$.
# Explanation:
## Step1: Set up equation to find $x$
Since $B$ is the mid - point of $\overline{AC}$, $AC = 2AB$. So $11x-17=2(3x + 4)$.
$$11x-17=6x + 8$$
Subtract $6x$ from both sides: $11x-6x-17=8$, so $5x-17=8$.
Add 17 to both sides: $5x=8 + 17 = 25$.
Divide both sides by 5: $x = 5$.
## Step2: Find $AC$
Substitute $x = 5$ into the expression for $AC$: $AC=11x - 17=11\times5-17=55-17 = 38$.
Since $AC = CD$, $CD = 38$.
## Step3: Find $DE$
$DE=CE - CD$, so $DE=49-38 = 11$.

### 6. If $m\angle DEF=117^{\circ}$, find the value of $x$.
Since $\angle DEF=(12x + 1)^{\circ}+(5x - 3)^{\circ}$, we have:
# Explanation:
## Step1: Set up equation for angle - sum
$(12x + 1)+(5x - 3)=117$.
$$12x+1+5x - 3=117$$
$$17x-2=117$$
## Step2: Solve for $x$
Add 2 to both sides: $17x=117 + 2=119$.
Divide both sides by 17: $x = 7$.

### 7. If $m\angle PQS = 16^{\circ}$, $m\angle SQR=(9x + 17)^{\circ}$, and $m\angle PQR=(12x - 6)^{\circ}$, find $m\angle PQR$.
# Explanation:
## Step1: Set up equation for angle - sum
Since $m\angle PQR=m\angle PQS+m\angle SQR$, we have $12x-6=16+(9x + 17)$.
$$12x-6=16+9x + 17$$
$$12x-6=9x + 33$$
## Step2: Solve for $x$
Subtract $9x$ from both sides: $12x-9x-6=33$, so $3x-6=33$.
Add 6 to both sides: $3x=33 + 6=39$.
Divide both sides by 3: $x = 13$.
## Step3: Find $m\angle PQR$
Substitute $x = 13$ into the expression for $m\angle PQR$: $m\angle PQR=12x - 6=12\times13-6=156-6 = 150^{\circ}$.

### 8. If $\overline{EF}$ bisects $\angle AED$, $m\angle AEF=(4x + 3)^{\circ}$, and $m\angle FED=(7x - 33)^{\circ}$, find $m\angle CEB$.
# Explanation:
## Step1: Set up equation based on angle - bisect property
Since $\overline{EF}$ bisects $\angle AED$, $m\angle AEF=m\angle FED$. So $4x + 3=7x - 33$.
## Step2: Solve for $x$
Subtract $4x$ from both sides: $3=7x-4x - 33$, so $3 = 3x-33$.
Add 33 to both sides: $3x=3 + 33=36$.
Divide both sides by 3: $x = 12$.
## Step3: Find $m\angle AED$
$m\angle AED=m\angle AEF+m\angle FED$. Substitute $x = 12$ into the expressions for $\angle AEF$ and $\angle FED$. $m\angle AEF=4\times12 + 3=48+3 = 51^{\circ}$, $m\angle FED=7\times12-33=84 - 33=51^{\circ}$, so $m\angle AED=51^{\circ}+51^{\circ}=102^{\circ}$. Since $\angle AED$ and $\angle CEB$ are vertical angles, $m\angle CEB=m\angle AED = 102^{\circ}$.

### 9. $\angle1$ and $\angle2$ form a linear pair. If $m\angle1=(18x - 1)^{\circ}$ and $m\angle2=(23x + 17)^{\circ}$, find $m\angle2$.
# Explanation:
## Step1: Set up equation based on linear - pair property
Since $\angle1$ and $\angle2$ form a linear pair, $m\angle1+m\angle2 = 180^{\circ}$. So $(18x - 1)+(23x + 17)=180$.
$$18x-1+23x + 17=180$$
$$41x+16=180$$
## Step2: Solve for $x$
Subtract 16 from both sides: $41x=180 - 16=164$.
Divide both sides by 41: $x = 4$.
## Step3: Find $m\angle2$
Substitute $x = 4$ into the expression for $m\angle2$: $m\angle2=23x + 17=23\times4+17=92 + 17=109^{\circ}$.

### 10. $\angle G$ and $\angle H$ are complementary angles. If $m\angle G=(6x - 15)^{\circ}$ and $m\angle H=(3x + 6)^{\circ}$, find $m\angle H$.
# Explanation:
## Step1: Set up equation based on complementary - angle property
Since $\angle G$ and $\angle H$ are complementary, $m\angle G+m\angle H = 90^{\circ}$. So $(6x - 15)+(3x + 6)=90$.
$$6x-15+3x + 6=90$$
$$9x-9=90$$
## Step2: Solve for $x$
Add 9 to both sides: $9x=90 + 9=99$.
Divide both sides by 9: $x = 11$.
## Step3: Find $m\angle H$
Substitute $x = 11$ into the expression for $m\angle H$: $m\angle H=3x + 6=3\times11+6=33 + 6=39^{\circ}$.

### 11. $\angle1$ and $\angle2$ are vertical angles. If $m\angle1=(5x + 12)^{\circ}$ and $m\angle2=(6x - 11)^{\circ}$, find $m\angle1$.
# Explanation:
## Step1: Set up equation based on vertical - angle property
Since $\angle1$ and $\angle2$ are vertical angles, $m\angle1=m\angle2$. So $5x + 12=6x - 11$.
## Step2: Solve for $x$
Subtract $5x$ from both sides: $12=6x-5x - 11$, so $x=12 + 11=23$.
## Step3: Find $m\angle1$
Substitute $x = 23$ into the expression for $m\angle1$: $m\angle1=5x + 12=5\times23+12=115 + 12=127^{\circ}$.

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QUESTION IMAGE

Question

Explanation:

1. If \(DF = 42\), find \(DE\).

Step1: Set up equation based on segment - addition

Step2: Solve for \(x\)

Step3: Find \(DE\)

2. In the diagram below, if \(JL=10x - 2\), \(JK = 5x - 8\), and \(KL=7x - 12\), find \(KL\).

Step1: Set up equation based on segment - addition

Step2: Solve for \(x\)

Step3: Find \(KL\)

3. If \(S\) is the mid - point of \(\overline{RT}\), \(RS = 5x+17\), and \(ST = 8x - 31\), find \(RS\).

Step1: Set up equation based on mid - point property

Step2: Solve for \(x\)

Step3: Find \(RS\)

4. If line \(y\) bisects \(\overline{AC}\), \(AB = 4-5x\), and \(BC=2x + 25\), find \(AC\).

Answer:

Step1: Set up equation based on vertical - angle property

Step2: Solve for \(x\)

Step3: Find \(m\angle1\)