Question

find the value of x.

1. m∠abc = 180°, m∠abd=(4x + 4)°
2. m∠qps=(3x - 15)°
3. m∠yxw = 168°, m∠wxz=(5x + 38)°
4. m∠lkj=(8x - 30)°
5. m∠utw=(7x + 15)°
6. m∠hef = 88°, m∠feg=(2x - 6)°

Explanation:

Step1: Set up equation for first problem

Since $\angle ABC = 180^{\circ}$ and $\angle ABD=(4x + 4)^{\circ}$, and $\angle DBC = 96^{\circ}$, then $(4x + 4)+96=180$. Simplify to get $4x+100 = 180$.

Step2: Solve for $x$ in first problem

Subtract 100 from both sides: $4x=180 - 100=80$. Then divide by 4: $x = 20$.

Step3: Set up equation for second problem

$\angle QPS=(3x - 15)^{\circ}$, and $\angle QPS=55 + 35=90^{\circ}$. So $3x-15 = 90$.

Step4: Solve for $x$ in second problem

Add 15 to both sides: $3x=90 + 15=105$. Divide by 3: $x = 35$.

Step5: Set up equation for third problem

$\angle YXW = 168^{\circ}$ and $\angle WXZ=(5x + 38)^{\circ}$, and $\angle ZXY = 100^{\circ}$. So $(5x + 38)+100=168$. Simplify to $5x+138 = 168$.

Step6: Solve for $x$ in third problem

Subtract 138 from both sides: $5x=168 - 138 = 30$. Divide by 5: $x = 6$.

Step7: Set up equation for fourth problem

$\angle LKJ=(8x - 30)^{\circ}$, and $\angle LKJ=23+27 = 50^{\circ}$. So $8x-30 = 50$.

Step8: Solve for $x$ in fourth problem

Add 30 to both sides: $8x=50 + 30=80$. Divide by 8: $x = 10$.

Step9: Set up equation for fifth problem

$\angle UTW=(7x + 15)^{\circ}$, and $\angle UTW=38+82 = 120^{\circ}$. So $7x+15 = 120$.

Step10: Solve for $x$ in fifth problem

Subtract 15 from both sides: $7x=120 - 15 = 105$. Divide by 7: $x = 15$.

Step11: Set up equation for sixth problem

$\angle HEF = 88^{\circ}$ and $\angle FEG=(2x - 6)^{\circ}$, and $\angle GEH = 40^{\circ}$. So $(2x - 6)+40=88$. Simplify to $2x+34 = 88$.

Step12: Solve for $x$ in sixth problem

Subtract 34 from both sides: $2x=88 - 34=54$. Divide by 2: $x = 27$.

Answer:

1. $x = 20$
2. $x = 35$
3. $x = 6$
4. $x = 10$
5. $x = 15$
6. $x = 27$