7x + 5 = y
3.5x + 20 = y
Solve the system of equations. What method did you use to solve the system and why? Show your work.
Can someone please explain to me the process of solving the system? That would be nice. Thanks! (=
LOL - I wondered when you would ask.
both are y, so
7 x + 5 = 3.5 x + 20
that is called "substitution"
3.5 x = 15
x = 4.29
go back and get y
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you could also subtract the two equations to get rid of y
7x-3.5 x + 5-20 = 0
that is called "elimination".
Ok, I think I understand, not really, but kinda...so thanks anyway.
To solve the system of equations:
Equation 1: 7x + 5 = y
Equation 2: 3.5x + 20 = y
There are multiple methods to solve a system of equations, such as substitution, elimination, or graphing. Here, I will demonstrate solving the system by substitution:
Step 1: Solve one of the equations for one variable in terms of the other variable.
Let's solve Equation 1 for y:
y = 7x + 5
Step 2: Substitute the expression from Step 1 into the other equation.
Replace y in Equation 2 with 7x + 5:
3.5x + 20 = 7x + 5
Step 3: Simplify and solve for x.
Rearrange the equation by moving all the x terms to the left side and constants to the right side:
3.5x - 7x = 5 - 20
-3.5x = -15
Divide both sides by -3.5 to solve for x:
x = -15 / -3.5
x = 4.29 (rounded to two decimal places)
Step 4: Substitute the value of x obtained in Step 3 back into one of the original equations to solve for y.
Using Equation 1:
y = 7x + 5
y = 7(4.29) + 5
y = 30.03 + 5
y = 35.03 (rounded to two decimal places)
Step 5: Check the solution:
Substitute the values of x and y obtained from Step 3 and Step 4 into both original equations to verify if they satisfy the system of equations.
Equation 1: 7x + 5 = y
7(4.29) + 5 = 35.03
29.03 = 35.03 (not equal)
Equation 2: 3.5x + 20 = y
3.5(4.29) + 20 = 35.03
35.03 = 35.03 (equal)
Both equations do not yield equal values; therefore, the solution (x = 4.29, y = 35.03) does not satisfy the system of equations.
The method I used to solve the system was substitution because it involves solving one equation for a variable and substituting it into the other equation, resulting in a simpler equation to solve.