Exercise 22 Page 204

Practice makes perfect

a We will use the Substitution Method to solve this system of equations. It is usually the best choice when one of the variables is already isolated or has a coefficient of 1 or -1. In the first equation y is already solved for, so we can substitute it in the second equation to find x.

y=3x+1 & (I) x+2y=-5 & (II)

Substitute

(II): y= 3x+1

y=3x+1 x+2( 3x+1)=-5

Distr

(II): Distribute 2

y=3x+1 x+6x+2=-5

▼

(II):Solve for x

AddTerms

(II): Add terms

y=3x+1 7x+2=-5

SubEqn

(II): LHS-2=RHS-2

y=3x+1 7x=-7

DivEqn

(II): .LHS /7.=.RHS /7.

y=3x+1 x=-1

Having found x, we can substitute this into the first equation to find y.

y=3x+1 x=-1

Substitute

(I): x= (-1)

y=3( -1)+1 x=-1

MultPosNeg

(I): a(- b)=- a * b

y=-3+1 x=-1

AddTerms

(I):Add terms

y=-2 x=-1

The solution to the system is (-1,-2).

b If we multiply the second equation by 2, the coefficient of the x-variables will be the same. This means we can use the Elimination Method to solve the system.

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

MultEqn

(II): LHS * 2=RHS* 2

2x+3y=9 2x-4y=2

SysEqnSub

(I): Subtract (II)

2x+3y-( 2x-4y)=9- 2 2x-4y=2

▼

(I):Solve for y

Distr

(I): Distribute (-1)

2x+3y-2x+4y=9-2 2x-4y=2