ANSWER
The solution is 
EXPLANATION
We want to solve the simultaneous equations
and
.
We substitute equation (2) in to equation (1), to obtain
This has now become a linear equation in a single variable 
.
We solve for x by grouping like terms.
We divide through by negative 3 to get;
.
Hence, the solution is
Simplifying
5x(4y + 3x) = 5x(3x + 4y)
Reorder the terms:
5x(3x + 4y) = 5x(3x + 4y)
(3x * 5x + 4y * 5x) = 5x(3x + 4y)
Reorder the terms:
(20xy + 15x2) = 5x(3x + 4y)
(20xy + 15x2) = 5x(3x + 4y)
20xy + 15x2 = (3x * 5x + 4y * 5x)
Reorder the terms:
20xy + 15x2 = (20xy + 15x2)
20xy + 15x2 = (20xy + 15x2)
Add '-20xy' to each side of the equation.
20xy + -20xy + 15x2 = 20xy + -20xy + 15x2
Combine like terms: 20xy + -20xy = 0
0 + 15x2 = 20xy + -20xy + 15x2
15x2 = 20xy + -20xy + 15x2
Combine like terms: 20xy + -20xy = 0
15x2 = 0 + 15x2
15x2 = 15x2
Add '-15x2' to each side of the equation.
15x2 + -15x2 = 15x2 + -15x2
Combine like terms: 15x2 + -15x2 = 0
0 = 15x2 + -15x2
Combine like terms: 15x2 + -15x2 = 0
0 = 0
Solving
0 = 0
Couldn't find a variable to solve for.
This equation is an identity, all real numbers are solutions.
Answer:
Step-by-step explanation:
You would simply change the x value s to its opposite and keep the y values the same.
