Exercise 9 Page 259

We are given a system of equations and asked to find the pair of

(x, y)

that satisfies both of the equations.

{3 x + 8 = 2 7 x + 3 y = 1 (I) (II)

Notice that in Equation (I) we have only

x -

terms, so we can start by finding the value of

x

using this equation.

{3 x + 8 = 2 7 x + 3 y = 1

$(I):$ $LHS - 8 = RHS - 8$

{3 x = - 6 7 x + 3 y = 1

$(I):$ $LHS / 3 = RHS / 3$

{x = - 2 7 x + 3 y = 1

Great! Now, to find the value of

y,

we need to substitute

x = - 2

into the second equation.

{x = - 2 7 x + 3 y = 1

$(II):$ $x = - 2$

{x = - 2 7 (- 2) + 3 y = 1

$(II):$ $a (- b) = - a \cdot b$

{x = - 2 - 14 + 3 y = 1

$(II):$ $LHS + 14 = RHS + 14$

{x = - 2 3 y = 15

$(II):$ $LHS / 3 = RHS / 3$

{x = - 2 y = 5

The solution, or point of intersection, to this system of equations is the point

(- 2, 5) .