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FLVS Algebra Two [Module Two]
FLVS Algebra Two [Module Two]
4
Mathematics
12th Grade
12/04/2010

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Cards

Term

Solve the system of equations by substitution.

x+y=3

x-2y=0

Definition

Solve for x in the first equation,

x=[3-y]

sub this value in for x in the second equation.

[3-y] -2y =0

3 -3y=0

-3y= -3

y= 1

Sub in to find x

x+1=3

x=2

Answer (2,1)

Term

Solve the system of equations by substitution.

3x - 2y = -4

2x + y = 9

 

Definition

solve for y in the second equation,

y=[9-2x]

sub this value in for y in the first equation.

3x - 2[9-2x] = -4

 3x-18+4x= -4

7x = 14

x=2

Sub in original equation to find y,

2[2] + y= 9

4 + y = 9

y = 5

Answer (2,5)

Term

Solve the system by graphing

3x - 2y ≥ 5

 x + 2y ≤ 7

Definition
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