Question

1. find the values of x and y. (10x - 61)° (18y + 5)° (x + 10)°

Explanation:

Step1: Set up equation for vertical - angles

Vertical - angles are equal. So, $10x−61=x + 10$.

Step2: Solve for x

Subtract x from both sides: $10x−x−61=x−x + 10$, which simplifies to $9x−61 = 10$. Then add 61 to both sides: $9x−61+61=10 + 61$, giving $9x=71$. Divide both sides by 9: $x=\frac{71}{9}\approx7.89$.

Step3: Set up equation for linear - pair

The angles $(18y + 5)$ and $(x + 10)$ form a linear - pair, so $(18y + 5)+(x + 10)=180$. Substitute $x=\frac{71}{9}$ into the equation: $18y+5+\frac{71}{9}+10 = 180$. First, combine the constant terms: $18y+(5 + 10)+\frac{71}{9}=180$, $18y + 15+\frac{71}{9}=180$. Convert 15 to a fraction with a denominator of 9: $15=\frac{135}{9}$. Then the equation becomes $18y+\frac{135}{9}+\frac{71}{9}=180$, $18y+\frac{135 + 71}{9}=180$, $18y+\frac{206}{9}=180$. Subtract $\frac{206}{9}$ from both sides: $18y=180-\frac{206}{9}$. Convert 180 to a fraction with a denominator of 9: $180=\frac{1620}{9}$. So $18y=\frac{1620}{9}-\frac{206}{9}=\frac{1620 - 206}{9}=\frac{1414}{9}$. Divide both sides by 18: $y=\frac{1414}{9\times18}=\frac{1414}{162}=\frac{707}{81}\approx8.73$.

Answer:

$x=\frac{71}{9},y=\frac{707}{81}$