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with No Solu+on and Iden+ty
Ac+vi+es by Jill
One Solu+on
3(2x + 4) = 6(5x + 2) Check
Iden+ty (Infinite Solu+ons)
3(x + 1) + 1 + 2x = 2(2x + 2) + x
Check
No Solu+on
8x + 3 -‐ 10x = -‐2(x -‐ 2) + 3
Check
Ac+vi+es by Jill 2013 Ac+vi+es by Jill
Ac+vi+es by Jill 2013
Ac+vi+es by Jill
Ac+vi+es by Jill 2013
I used this foldable to introduce equa+ons
with infinitely many
solu+ons and no solu+on.
Ac+vi+es by Jill
Mul+-‐Step Equa+ons Name __________________
1) 2) 3)
4) 5) 6)
Solve each equa+on. Show all work clearly and circle your final answer. Equa+ons may have one solu+on, no solu+on, or be an iden+ty with infinite solu+ons.
Objec+ve: Students will solve mul+-‐step equa+ons having one solu+on, no solu+on, or infinite solu+ons.
−12x+4( )=18 −42=− 3
4f−4( ) 3 z−5( )+17=−4
1315−6p( )= 4−2p 9−2y
3= y 3 w+5( )−6=3(3+w)
Ac+vi+es by Jill
7) 8) 9)
10) 11) 12)
Objec+ve: Students will solve mul+-‐step equa+ons having one solu+on, no solu+on, or infinite solu+ons.
7− 2n+1( )=8n 6g−2 2−g( )= 4 2g−1( ) 3+4 t+2( )=2t−3 t+4( )
5k+3 1−2 k+1( )"#
$%=2k 3 j+2( )− j=2 j+1( ) 8+e= 1
5e+ 4
5e−10( )
Check: If you have solved all twelve equa+ons correctly, there should be 3 with no solu+on, 2 iden++es, and the sum of those with one solu+on should be 14.8.
Ac+vi+es by Jill
Mul+-‐Step Equa+ons Name __________________
1) 2) 3)
4) 5) 6)
Solve each equa+on. Show all work clearly and circle your final answer. Equa+ons may have one solu+on, no solu+on, or be an iden+ty with infinite solu+ons.
Objec+ve: Students will solve mul+-‐step equa+ons having one solu+on, no solu+on, or infinite solu+ons.
−12x+4( )=18
−12x−2=18
−12x =20
x =−40
−42=− 34
f−4( )−42=− 3
4f+3
−45=− 34f
f =60
3 z−5( )+17=−43z−15+17=−43z+2=−43z=−6z=−2
1315−6p( )= 4−2p
5−2p= 4−2p5≠ 4
9−2y3
= y
9−2y =3y9=5y
y = 95=14
5=1.8
3 w+5( )−6=3(3+w)3w+15−6=9+3w3w+9=9+3w9=9
No Solu+on Iden+ty
Answer Key
Ac+vi+es by Jill
7) 8) 9)
10) 11) 12)
Objec+ve: Students will solve mul+-‐step equa+ons having one solu+on, no solu+on, or infinite solu+ons.
Check: If you have solved all twelve equa+ons correctly, there should be 3 with no solu+on, 2 iden++es, and the sum of those with one solu+on should be 14.8.
7− 2n+1( )=8n7−2n−1=8n−2n+6=8n6=10n
n= 610
=35= 0.6
6g−2 2−g( )= 4 2g−1( )6g−4+2g =8g−48g−4 =8g−4−4 =−4
3+4 t+2( )=2t−3 t+4( )3+4t+8=2t−3t−124t+11=−t−125t+11=−125t =−23
t =−235=−43
5=−4.6
5k+3 1−2 k+1( )"#
$%=2k
5k+3 1−2k−2( )=2k5k+3 −2k−1( )=2k5k−6k−3=2k−k−3=2k−3=3kk =−1
3 j+2( )− j=2 j+1( )3j+6− j=2j+22j+6=2j+26≠2
8+e= 15e+ 4
5e−10( )
8+e= 15e+ 4
5e−8
8+e=e−88≠−8
Iden+ty
No Solu+on
No Solu+on
Ac+vi+es by Jill
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Ac+vi+es by Jill