Solve the system of equations.%0D%0A6x–5y=27%0D%0A6%0D%0A%0D%0A–%0D%0A5%0D%0A%0D%0A=%0D%0A27%0D%0A%0D%0A3x+5y=36%0D%0A3%0D%0A%0D%0A+%0D%0A5%0D%0A%0D%0A=%0D%0A36%0D%0A(1 point)%0D%0A(, )
To solve the system of equations, we can use the method of substitution or elimination.
Using substitution:
From the first equation, we can isolate x:
6x - 5y = 27
6x = 5y + 27
x = (5y + 27)/6
Substituting this value of x into the second equation:
3(5y + 27)/6 + 5y = 36
Multiply both sides by 6 to eliminate the denominator:
3(5y + 27) + 30y = 216
15y + 81 + 30y = 216
45y + 81 = 216
45y = 135
y = 135/45
y = 3
Substituting y = 3 back into the first equation to find x:
x = (5(3) + 27)/6
x = (15 + 27)/6
x = 42/6
x = 7
Therefore, the solution to the system of equations is (x, y) = (7, 3).
To solve the system of equations, we can use the method of elimination. The goal is to eliminate one variable by adding or subtracting the equations.
Let's start by multiplying the entire second equation by 2 to make the coefficients of y in both equations the same:
2(3x + 5y) = 2(36)
6x + 10y = 72
Now we can subtract the first equation from the second equation to eliminate the x variable:
(6x + 10y) - (6x - 5y) = 72 - 27
6x - 6x + 10y + 5y = 45
15y = 45
Divide both sides by 15:
y = 45/15
y = 3
Now substitute the value of y back into one of the original equations, let's use the first equation:
6x - 5(3) = 27
6x - 15 = 27
Add 15 to both sides:
6x = 42
Divide both sides by 6:
x = 42/6
x = 7
Therefore, the solution to the system of equations is x = 7 and y = 3.
To solve the system of equations:
1) First, we write the equations in standard form (Ax + By = C):
Equation 1: 6x − 5y = 27
Equation 2: 3x + 5y = 36
2) To eliminate one of the variables, we can add the two equations together. By doing this, the y terms will cancel out:
(6x - 5y) + (3x + 5y) = 27 + 36
9x = 63
3) Now we solve for x by dividing both sides of the equation by 9:
x = 63/9
x = 7
4) Substitute the value of x back into one of the original equations. Let's use Equation 1:
6(7) − 5y = 27
42 - 5y = 27
5) Solve for y by subtracting 42 from both sides of the equation:
-5y = 27 - 42
-5y = -15
6) Divide both sides of the equation by -5:
y = -15 / -5
y = 3
7) The solution to the system of equations is x = 7, y = 3.
Therefore, the solution to the system of equations is (7, 3).