Presentation is loading. Please wait.

Presentation is loading. Please wait.

Solving linear equations: Variable on both sides and brackets

Similar presentations


Presentation on theme: "Solving linear equations: Variable on both sides and brackets"β€” Presentation transcript:

1 Solving linear equations: Variable on both sides and brackets
Silent Teacher Intelligent Practice Narration Your Turn 2(π‘₯+3)=π‘₯+5 3(π‘₯+3)=2(π‘₯+5) Practice

2 Worked Example Your Turn 2(π‘₯+1)=π‘₯+5 3(π‘₯+1)=π‘₯+5 @mathsmiles_k

3 π‘₯+5=3 π‘₯+1 3 π‘₯βˆ’5 =3(2π‘₯+1) 3 π‘₯+5 =π‘₯+1 3 π‘₯βˆ’5 =βˆ’3(2π‘₯+1) 3 π‘₯+5 =3(2π‘₯+1)
1. 7. 3 π‘₯βˆ’5 =3(2π‘₯+1) 3 π‘₯+5 =π‘₯+1 2. 8. 3 π‘₯βˆ’5 =βˆ’3(2π‘₯+1) 3. 3 π‘₯+5 =3(2π‘₯+1) 9. βˆ’3 π‘₯+5 =βˆ’3(2π‘₯+1) 4. 3 π‘₯+5 =3(2π‘₯βˆ’1) 10. βˆ’3 π‘₯βˆ’5 =βˆ’3(2π‘₯+1) 5. 3 π‘₯βˆ’5 =3(2π‘₯βˆ’1) 11. βˆ’3 π‘₯βˆ’5 =βˆ’3(2π‘₯βˆ’1) 6. 3 2π‘₯βˆ’5 =3(π‘₯βˆ’1) 12. βˆ’3 2π‘₯βˆ’1 =βˆ’3(π‘₯βˆ’5) @mathsmiles_k

4 π‘₯+5=3 π‘₯+1 3 π‘₯βˆ’5 =βˆ’3(2π‘₯+1) π‘₯=1 π‘₯=βˆ’6 3 π‘₯+5 =π‘₯+1 βˆ’3 π‘₯+5 =βˆ’3(2π‘₯+1) π‘₯=-7
π‘₯= 12 9 3 π‘₯+5 =3(2π‘₯+1) π‘₯=4 βˆ’3 π‘₯βˆ’5 =βˆ’3(2π‘₯+1) π‘₯=4 3 π‘₯+5 =3(2π‘₯βˆ’1) π‘₯=6 βˆ’3 π‘₯βˆ’5 =βˆ’3(2π‘₯βˆ’1) 3 π‘₯βˆ’5 =3(2π‘₯βˆ’1) π‘₯=βˆ’4 π‘₯=βˆ’4 βˆ’3 2π‘₯βˆ’1 =βˆ’3(π‘₯βˆ’5) π‘₯=βˆ’4 3 2π‘₯βˆ’5 =3(π‘₯βˆ’1) π‘₯=4 @mathsmiles_k


Download ppt "Solving linear equations: Variable on both sides and brackets"

Similar presentations


Ads by Google