Question

unit 5, lesson 3 – ready, set, go
ready
identify which of the given points are solutions to the following linear equations. there are multiple solutions for each equation.

1. $3x + 2y = 12$

a. $(2,4)$
b. $(3,2)$
c. $(4,0)$
d. $(0,6)$

1. $5x - y = 10$

a. $(2,0)$
b. $(3,0)$
c. $(0,-10)$
d. $(1,1)$

1. $-x + 6y = 10$

a. $(-4,1)$
b. $(-22,-2)$
c. $(2,2)$
d. $(10,0)$

Explanation:

Step1: Check Eq1: $3x+2y=12$

Test A. (2,4):

Substitute $x=2,y=4$:
$3(2)+2(4)=6+8=14
eq12$

Test B. (3,2):

Substitute $x=3,y=2$:
$3(3)+2(2)=9+4=13
eq12$

Test C. (4,0):

Substitute $x=4,y=0$:
$3(4)+2(0)=12+0=12$

Test D. (0,6):

Substitute $x=0,y=6$:
$3(0)+2(6)=0+12=12$

Step2: Check Eq2: $5x-y=10$

Test A. (2,0):

Substitute $x=2,y=0$:
$5(2)-0=10-0=10$

Test B. (3,0):

Substitute $x=3,y=0$:
$5(3)-0=15-0=15
eq10$

Test C. (0,-10):

Substitute $x=0,y=-10$:
$5(0)-(-10)=0+10=10$

Test D. (1,1):

Substitute $x=1,y=1$:
$5(1)-1=5-1=4
eq10$

Step3: Check Eq3: $-x+6y=10$

Test A. (-4,1):

Substitute $x=-4,y=1$:
$-(-4)+6(1)=4+6=10$

Test B. (-22,-2):

Substitute $x=-22,y=-2$:
$-(-22)+6(-2)=22-12=10$

Test C. (2,2):

Substitute $x=2,y=2$:
$-(2)+6(2)=-2+12=10$

Test D. (10,0):

Substitute $x=10,y=0$:
$-(10)+6(0)=-10+0=-10
eq10$

Answer:

1. C. (4, 0), D. (0, 6)
2. A. (2, 0), C. (0, - 10)
3. A. (- 4, 1), B. (- 22, - 2), C. (2, 2)