1st Change to y = mx + b •x + 2y = 7 •x + y = 1
•x-x+2y=-x+7 •x-x+y=-x+1 •y = -x + 1 •2y=-x+7 •2y/2=-x/2+7/2 •y= -1/2x + 7/2
y = -1/2x + 7/2 y = -x + 1
7/2 = 3 1/2
(-5,6)
Notice the intersection is not at 2 pts.
Objective
•To be able to solve a system of equations by substitution….
1.) Solve by Substitution
• y = 3x - 1 • 2x + y = 9 So…... • 2x + (3x-1) = 9 • 5x - 1 = 9 (add 2x and 3x) • 5x = 10 (add 1 to both sides) • x = 2 (divide both by 5)
1.) Solve by Substitution
• Since x = 2 We can now substitute it back in • y = 3x - 1 (the 1st equation) • y = 3(2) -1(put in the x value) • y = 6 - 1 (solve for y) •y = 5 (2,5) is the solution
1.) Solve by Substitution
• Or …. x = 2 We can substitute it in the other one • 2x + y = 9 (the 2nd equation) • 2(2) + y = 9(put in the x value) • 4 + y = 9(subtract 4 from both) •y = 5 (solve for y)
Graph y = 3x -1 & 2x + y = 9
2x+y=9
y = 3x - 1
Solution= (2,5)
2.) Solve by Substitution
• 2x + 3y = 7 • y + 1 = x So….. • 2(y+1) + 3y = 7 • 2y +2 + 3y = 7(distribute the 2) • 5y + 2 = 7(Add 2y & 3y) • 5y = 5 & y = 1(divide both by5)
2.) Solve by Substitution
• Since y = 1 We can now substitute it back in • y + 1 = x (the 2nd equation) • 1 + 1 = x (put in the y value) • 2 = x (solve for x) • (2,1) is the solution
Solve by Substitution
• y = 3x You try • y + 4x = 28 this one • First find what x or y equals • Then Substitute into the other equation.
Solve by Substitution
• y = 3x • y + 4x = 28 • 3x + 4x = 28 • 7x = 28 •x = 4 • Since x = 4 • y = 3x • y = 3(4) • y = 12 • solution (4,12)
3.) Solve by Substitution
• 2x + 4y = 8 First solve • 3x + 5y = 14 for x or y • 1st… 2x = -4y + 8 (Subtract 4y) x = -2y + 4 (Divide by 2) • Now substitute for x • 3(-2y + 4) + 5y = 14
3.) Solve by Substitution
•3(-2y + 4) + 5y = 14 •-6y + 12 + 5y = 14 •-y + 12 = 14 •-y = 2 so…. y = -2
3.) Solve by Substitution
• Since y = -2 and x = -2y + 4 • Then x = -2(-2) + 4 •x = 4 + 4 •x = 8 • Solution (8,-2)