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McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

3. Elimination Using Addition and Subtraction

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Exercise 46 Page 356

Does either of the equations have an isolated variable in it?

$(- 14, - 4)$

Practice makes perfect

When solving a system of equations using substitution, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

For this exercise,

x

is already isolated in both equations, so we can skip straight to solving!

{x = 5 y + 6 x = 3 y - 2 (I) (II)

$(II):$ $x = 5 y + 6$

{x = 5 y + 6 5 y + 6 = 3 y - 2

$(II):$ $LHS - 6 = RHS - 6$

{x = 5 y + 6 5 y = 3 y - 8

$(II):$ $LHS - 3 y = RHS - 3 y$

{x = 5 y + 6 2 y = - 8

$(II):$ $LHS / 2 = RHS / 2$

{x = 5 y + 6 y = - 4

Great! Now, to find the value of

x,

let's substitute

- 4

for

y

in Equation (I).

{x = 5 y + 6 y = - 4

$(I):$ $y = - 4$

{x = 5 (- 4) + 6 y = - 4

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

{x = - 20 + 6 y = - 4

$(I):$ Add terms

{x = - 14 y = - 4

The solution to this system of equations is

(- 14, - 4) .

Subchapter links

Check Your Understanding p.353

H.O.T. Problems p.355

Standardized Test Practice p.356

Spiral Review p.356

Skills Review p.356