I need help solving this set of equations by substitution.
xy=2
3x − y + 5 = 0
I originally used (3x+5=y) and came up with (x=.333 and x=-2) but when I substitute it into the original equation I get (y=2 and y=1) and it is wrong according to my homework. Please help me!
ok, go with y = 3x+5
into the 1st
x(3x+5) = 2
3x^2 + 5x - 2 = 0
((3x-1)(x+2) = 0
x = 1/3 or x = -2 --- > you are correct
if x = 1/3, y= 3(1/3) + 5 = 6
if x = -2 , y = 3(-2) + 5 = -1
so x=1/3, y=6 OR x=-2, y =-1
I don't know how you got your values ???
To solve this set of equations by substitution, you first need to isolate one variable in one equation and substitute it into the other equation.
Let's start with the first equation: xy = 2.
Since the goal is to isolate one variable, we can isolate y by dividing both sides of the equation by x:
y = 2/x.
Now, we substitute this expression for y into the second equation:
3x - (2/x) + 5 = 0.
To simplify this equation, we can multiply through by the denominator x to get rid of the fraction:
3x^2 - 2 + 5x = 0.
Now we have a quadratic equation. To solve for x, we can rearrange the terms and set the equation equal to zero:
3x^2 + 5x - 2 = 0.
Next, you can either factor, complete the square, or use the quadratic formula to solve for x. In this case, the quadratic equation can be factored:
(3x - 1)(x + 2) = 0.
This gives us two possible solutions:
1) 3x - 1 = 0, which gives x = 1/3.
2) x + 2 = 0, which gives x = -2.
Now that we have the possible values for x, we can substitute them back into the original equation to find the corresponding y-values.
For x = 1/3:
y = 2/(1/3) = 6.
For x = -2:
y = 2/(-2) = -1.
So the solutions to the system of equations are:
x = 1/3, y = 6 (Solution 1)
x = -2, y = -1 (Solution 2)
Make sure to double-check your calculations and ensure that you've correctly substituted the values back into the original equations.