1 / 14

Solving Systems of Equations By Substitution – Harder

Learn how to solve a system of equations using the substitution method. Follow step-by-step instructions to find the values of the variables and check your solutions. Examples provided.

helenholmes
Download Presentation

Solving Systems of Equations By Substitution – Harder

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Dr. Fowler  CCM Solving Systems of EquationsBy Substitution – Harder

  2. Solving a system of equations by substitution Pick the easier equation. The goal is to get y= ; x= ; a= ; etc. Step 1: Solve an equation for one variable. Step 2: Substitute Put the equation solved in Step 1 into the other equation. Step 3: Solve the equation. Get the variable by itself. Step 4: Plug back in to find the other variable. Substitute the value of the variable into the equation. Step 5: Check your solution. Substitute your ordered pair into BOTH equations. ALREADY IN NOTES – Read Only for Review

  3. 1) Solve the system using substitution 3y + x = 7 4x – 2y = 0 It is easiest to solve the first equation for x. 3y + x = 7 -3y -3y x = -3y + 7 Step 1: Solve an equation for one variable. Step 2: Substitute 4x – 2y = 0 4(-3y + 7) – 2y = 0

  4. 1) Solve the system using substitution 3y + x = 7 4x – 2y = 0 -12y + 28 – 2y = 0 -14y + 28 = 0 -14y = -28 y = 2 Step 3: Solve the equation. 4x – 2y = 0 4x – 2(2) = 0 4x – 4 = 0 4x = 4 x = 1 Step 4: Plug back in to find the other variable.

  5. 1) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 5: Check your solution. (1, 2) 3(2) + (1) = 7 4(1) – 2(2) = 0 Answer is (1,2)

  6. 2) Solve the following system using the substitution method. 3x – y = 6 and – 4x + 2y = –8 STEP 1 – Solve the first equation for y is easiest, 3x – y = 6 –y = –3x + 6 (subtract 3x from both sides) y = 3x – 6 (multiply both sides by – 1) STEP 2 – Substitute this value for y in the OTHER equation. –4x + 2y = –8 –4x + 2(3x – 6) = –8 (replace y to other equation) –4x + 6x – 12 = –8 (use the distributive property) 2x – 12 = –8 (simplify the left side) 2x = 4 (add 12 to both sides) x = 2 (divide both sides by 2) CONTINUED >

  7. STEP 3 – To get y, substitute x = 2 into either original equation. The easiest is the one that was already solved for y. y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0y = 0 We have now found x & y. Answer is (2, 0)

  8. EXAMPLE 3 • Solve the system by the substitution method. Solve first for x: Substitute: TRUE – The answer is infinitely many solutions

  9. EXAMPLE 4 • Use substitution to solve the system. Solve first for x: Substitute into other equation: To get X, substitute y you found into equation already solved for X:

  10. Example #5: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. x + y = 10 y = -x +10 Step 2: Substitute into the other equation. 5x - y = 2 5x -(-x +10) = 2

  11. x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x = 2

  12. x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8).

  13. Excellent Job !!!Well Done

  14. Stop Notes Do Worksheet

More Related