Solve Systems By Substitution (Harder Version)

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Systems of Equations

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

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

Slide 3  2) 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
 When is solving systems by substitution easier to do than graphing?
 When only one of the equations has a variable already isolated (like in example #1).

Slide 4  If you solved the first equation for x, what would be substituted into the bottom equation.
 2x + 4y = 4
 3x + 2y = 22
 4y + 4
 2y + 2
 2x + 4
 2y+ 22

Slide 6  3) Solve the system using substitution
 2x + y = 4
 4x + 2y = 8
 Step 1: Solve an equation for one variable.
 Step 2: Substitute
 The first equation is
 easiest to solved for y!
 y = 2x + 4
 4x + 2y = 8
 4x + 2(2x + 4) = 8
 Step 3: Solve the equation.
 4x – 4x + 8 = 8
 8 = 8
 This is also a special case.
 Does 8 = 8? TRUE!
 When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS.

Slide 7  What does it mean if the result is “TRUE”?