how do i solve the equation for x and y
y+4x=27
xy+x=40
y+4x=27
y = 27-4x
Substitute 27-4x for y in the second equation to find value for x. Put this value in the above equation to find the value for y. Put both values in the second equation to check your answers.
I hope this helps. Thanks for asking.
To solve the system of equations for x and y, you can use either the substitution method or the elimination method. Let's solve it using the substitution method:
1. Solve the first equation for y:
y + 4x = 27
y = 27 - 4x
2. Substitute the value of y in the second equation with 27 - 4x:
xy + x = 40
(27 - 4x)x + x = 40
3. Simplify the equation:
27x - 4x^2 + x = 40
4. Rearrange the equation in standard form:
-4x^2 + 28x - 40 = 0
5. Divide the equation by -4 to simplify:
x^2 - 7x + 10 = 0
6. Factorize the quadratic equation:
(x - 5)(x - 2) = 0
7. Set each factor equal to zero to find the values of x:
x - 5 = 0 --> x = 5
x - 2 = 0 --> x = 2
8. Substitute the values of x back into one of the original equations to solve for y:
For x = 5:
y + 4(5) = 27
y + 20 = 27
y = 7
For x = 2:
y + 4(2) = 27
y + 8 = 27
y = 19
Therefore, the solutions to the system of equations are:
x = 5, y = 7
or
x = 2, y = 19
To solve the system of equations for x and y, you can use either substitution method or elimination method. Let's use the elimination method in this case:
First, let's rearrange the equations to isolate the variables:
Equation 1: y + 4x = 27
Equation 2: xy + x = 40
To eliminate the y-term, multiply both sides of Equation 1 by x:
Equation 3: xy + 4x^2 = 27x
Now, subtract Equation 3 from Equation 2:
Equation 2 - Equation 3: (xy + x) - (xy + 4x^2) = 40 - 27x
Simplifying, we get:
x - 4x^2 = 40 - 27x
Rearrange the terms to form a quadratic equation:
-4x^2 - 27x + x - 40 = 0
Combine like terms:
-4x^2 - 26x - 40 = 0
To solve quadratic equations, you can use either factoring, completing the square, or the quadratic formula. In this case, the equation cannot be easily factored, so let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our quadratic equation, a = -4, b = -26, and c = -40.
Substitute the values into the quadratic formula:
x = (-(-26) ± sqrt((-26)^2 - 4(-4)(-40))) / (2(-4))
Simplifying:
x = (26 ± sqrt(676 - 640)) / (-8)
x = (26 ± sqrt(36)) / (-8)
x = (26 ± 6) / (-8)
Now, we have two possible values for x:
x1 = (26 + 6) / (-8) = 32 / (-8) = -4
x2 = (26 - 6) / (-8) = 20 / (-8) = -2.5
Substitute each value of x back into either equation to find the corresponding values of y.
Using Equation 1:
When x = -4:
y + 4(-4) = 27
y - 16 = 27
y = 27 + 16
y = 43
When x = -2.5:
y + 4(-2.5) = 27
y - 10 = 27
y = 27 + 10
y = 37
So, the solutions to the system of equations are:
(x, y) = (-4, 43) and (-2.5, 37)