We can solve this system of equations using either the substitution method or the elimination method.
Substitution Method:
1. Solve one equation for one variable in terms of the other variable. Let's solve the first equation for x:
3x = 40 + 4y
x = (40 + 4y)/3
2. Substitute this expression for x in the second equation:
5((40 + 4y)/3) - 5y = 55
3. Simplify and solve for y:
(200 + 20y)/3 - 5y = 55
Multiply both sides of the equation by 3 to eliminate the fraction:
200 + 20y - 15y = 165
5y = -35
y = -7
4. Substitute this value of y back into one of the original equations and solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
The solution to the system of equations is x = 4 and y = -7.
Elimination Method:
1. Multiply the first equation by 5 and the second equation by 3 to make the coefficients of x in both equations the same:
15x - 20y = 200
15x - 15y = 165
2. Subtract the second equation from the first equation to eliminate the x terms:
(15x - 20y) - (15x - 15y) = 200 - 165
5y = -35
y = -7
3. Substitute this value of y back into one of the original equations and solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
The solution to the system of equations is x = 4 and y = -7.
Solve the system of equations.
3x - 4y = 40
5x - 5y = 55 (1 point)
1 answer