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A straight line has the equation $ax + by = 5$ , where $a$ and $b$ are constants. It passes through the points $(3, 2)$ and $(1, -1)$ . Find (a) the value of $a$ and of $b$ , (b) the gradient of the line.

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Write the equation of a linear function

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Q. A straight line has the equation

ax + by = 5

, where

a

and

b

are constants. It passes through the points

(3, 2)

and

(1, -1)

. Find (a) the value of

a

and of

b

, (b) the gradient of the line.

Plug Coordinates Equation: Plug in the coordinates of the first point $(3, 2)$ into the equation $ax + by = 5$ . $\newline$ $3a + 2b = 5$
Plug Second Point Equation: Plug in the coordinates of the second point $(1, -1)$ into the equation $ax + by = 5$ . $\newline$ $a - b = 5$
Solve System Equations: Now we have a system of equations: $\newline$ $3a + 2b = 5$ $\newline$ $a - b = 5$ $\newline$ Let's solve this system by multiplying the second equation by $2$ to eliminate $b$ . $\newline$ $2(a - b) = 2(5)$ $\newline$ $2a - 2b = 10$
Eliminate Variable: Add the new equation $2a - 2b = 10$ to the first equation $3a + 2b = 5$ to eliminate $b$ . $\newline$ $3a + 2b + 2a - 2b = 5 + 10$ $\newline$ $5a = 15$
Find Value of a: Divide both sides by $5$ to find the value of a. $\newline$ $a = \frac{15}{5}$ $\newline$ $a = 3$
Substitute $a$ Find $b$ : Substitute $a = 3$ back into the second original equation $a - b = 5$ to find $b$ . $\newline$ $3 - b = 5$
Solve for b: Subtract $3$ from both sides to solve for $b$ .
$-b = 5 - 3$
$-b = 2$
$b = -2$
Calculate Gradient: Now, calculate the gradient (slope) of the line using the two points $(3, 2)$ and $(1, -1)$ . $\newline$ Gradient $(m) = \frac{y_2 - y_1}{x_2 - x_1}$ $\newline$ $m = \frac{-1 - 2}{1 - 3}$ $\newline$ $m = \frac{-3}{-2}$ $\newline$ $m = \frac{3}{2}$

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