body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

3. Elimination Using Addition and Subtraction

Continue to next subchapter

Exercise 43 Page 356

In this system of equations, at least one of the variables has a coefficient of $1 .$ Therefore, we will approach its solution with the Substitution Method. 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.

Observing the given equations, it looks like it will be simplest to isolate

x

in the first equation.

{x + 4 y = 1 2 x - 3 y = - 9 (I) (II)

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

{x = 1 - 4 y 2 x - 3 y = - 9

Now that we've isolated

x,

we can solve the system by substitution.

{x = 1 - 4 y 2 x - 3 y = - 9

$(II):$ $x = 1 - 4 y$

{x = 1 - 4 y 2 (1 - 4 y) - 3 y = - 9

$(II):$ Distribute $2$

{x = 1 - 4 y 2 - 8 y - 3 y = - 9

$(II):$ Subtract term

{x = 1 - 4 y 2 - 11 y = - 9

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

{x = 1 - 4 y - 11 y = - 11

$(II):$ $LHS / (- 11) = RHS / (- 11)$

{x = 1 - 4 y y = 1

Now, let's substitute

1

for

y

in the first equation to find the value of

x .

{x = 1 - 4 y y = 1

$(I):$ $y = 1$

{x = 1 - 4 (1) y = 1

$(I):$ Multiply

{x = 1 - 4 y = 1

$(I):$ Subtract term

{x = - 3 y = 1

The solution to this system of equations is the point

(- 3, 1) .

This corresponds to answer B.

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