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Solve the system of equation using substitution 2x plus 5y equals 2 and -3x-y equals -3?
2x + 5y = 2-3x - y = -3from the second equation:-y = -3 + 3xy = 3 - 3xsubstitute this value in the first equation:2x + 5(3-3x) = 22x + 15 - 15x = 2-13x + 15 = 2-13x = 2 - 15-13x = -13so, x = 1Now, solve for y :y = 3 - 3xy = 3 - 3 (1)y = zero
Solve for y -3x plus 6y equals 12?
y-3x+6y=12 7y=3x+12 y=3x/7 +12/7
How do you solve this equation 3x plus 2y equals 12 solve for y?
3x + 2y = 12 ie 2y = 12 - 3x so y = 6 - 3x/2
What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8 showing all appropiate work with answers?
If: x-2y = 1 and 3xy -y^2 = 8 Then: x = 1+2y and 3(1+2y)y -y^2 = 8 Hence: 3y+6y^2 -y^2 = 8 => 5y^2 +3y -8 = 0 Solving the above quadratic equation: y =1 or y = -8/5 Solutions by substitution are: when y=1 then x=3 and when y=-8/5 then x=-11/5
How do you solve the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8?
x - 2y = 1 → x = 1 + 2y 3xy - y² = 8 Substitute the first equation into the second equation and subtract 8 from both sides: 3(1 +2y)y - y² = 0 3y +6y² - y² - 8 = 0 5y² + 3y - 8 = 0 When factored: (5y +8)(y -1) = 0 Therefore: y = -⁸/₅ or y = 1 Substituting the above values into the linear equation gives the solutions as: (3, 1) and (-¹¹/₈, -⁸/₅)
