Question

in exercises 25 - 28, find the missing coefficient in the equation of the line shown. write the completed equation. (see example 4.) 25. ax + 3y = 5 26. ax - 4y = -1 27. -x + by = 10 28. 8x + by = 4

Explanation:

Step1: Substitute point into equation

For exercise 25, the equation is $Ax + 3y=5$ and the point $(2,-1)$ lies on the line. Substitute $x = 2$ and $y=-1$ into the equation: $A\times2+3\times(-1)=5$.

Step2: Solve for $A$

Simplify the left - hand side of the equation: $2A-3 = 5$. Add 3 to both sides: $2A=5 + 3=8$. Divide both sides by 2: $A = 4$.

For exercise 26, the equation is $Ax-4y=-1$ and the point $(6,1)$ lies on the line. Substitute $x = 6$ and $y = 1$ into the equation: $A\times6-4\times1=-1$.

Step3: Solve for $A$

Simplify the left - hand side: $6A-4=-1$. Add 4 to both sides: $6A=-1 + 4=3$. Divide both sides by 6: $A=\frac{1}{2}$.

For exercise 27, the equation is $-x+By = 10$ and the point $(-2,2)$ lies on the line. Substitute $x=-2$ and $y = 2$ into the equation: $-(-2)+B\times2=10$.

Step4: Solve for $B$

Simplify the left - hand side: $2 + 2B=10$. Subtract 2 from both sides: $2B=10 - 2=8$. Divide both sides by 2: $B = 4$.

For exercise 28, the equation is $8x+By = 4$ and the point $(-5,4)$ lies on the line. Substitute $x=-5$ and $y = 4$ into the equation: $8\times(-5)+B\times4=4$.

Step5: Solve for $B$

Simplify the left - hand side: $-40+4B=4$. Add 40 to both sides: $4B=4 + 40=44$. Divide both sides by 4: $B = 11$.

Answer:

Exercise 25: $A = 4$
Exercise 26: $A=\frac{1}{2}$
Exercise 27: $B = 4$
Exercise 28: $B = 11$