Solve the system by substitution. Check your solution.
[tex]a-12 b=-3[/tex]
[tex]0.2 b+0.6 a=12[/tex]
A. [tex](15,15)[/tex]
B. [tex](10,12)[/tex]
C. [tex](13,12)[/tex]
D. [tex](7,9)[/tex]
Answer (2)
Solve the first equation for a : a = 12 b − 3 .
Substitute this expression into the second equation: 0.2 b + 0.6 ( 12 b − 3 ) = 12 .
Simplify and solve for b : b = 37 69 .
Substitute b back into the equation for a : a = 37 717 . The correct answer is not among the given choices.
Explanation
Analyze the problem We are given the following system of equations:
a − 12 b = − 3 0.2 b + 0.6 a = 12
We will solve this system using the substitution method.
Solve for a First, solve the first equation for a in terms of b :
a = 12 b − 3
Substitute into second equation Substitute this expression for a into the second equation:
0.2 b + 0.6 ( 12 b − 3 ) = 12
Solve for b Simplify and solve for b :
0.2 b + 7.2 b − 1.8 = 12 7.4 b = 13.8 b = 7.4 13.8 = 74 138 = 37 69 ≈ 1.86486
Solve for a Substitute the value of b back into the expression for a :
a = 12 ( 37 69 ) − 3 = 37 828 − 37 111 = 37 717 ≈ 19.37838
Compare to given options The solution to the system of equations is approximately a = 19.37838 and b = 1.86486 . Now we compare this solution to the given options. None of the options match our solution.
Double check the solution Let's re-examine the steps to make sure there were no mistakes. The equations are:
a − 12 b = − 3 0.2 b + 0.6 a = 12
Solving the first equation for a :
a = 12 b − 3
Substituting into the second equation:
0.2 b + 0.6 ( 12 b − 3 ) = 12 0.2 b + 7.2 b − 1.8 = 12 7.4 b = 13.8 b = 7.4 13.8 = 37 69
Substituting back into the equation for a :
a = 12 ( 37 69 ) − 3 = 37 828 − 37 111 = 37 717
So the solution is a = 37 717 and b = 37 69 .
Final Answer Since none of the provided options match the correct solution, we can conclude that there is no correct answer among the given choices.
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business. For example, a company might want to know how many units they need to sell to cover their costs. This involves setting up equations for cost and revenue and solving for the quantity at which they are equal. Similarly, in finance, systems of equations can be used to optimize investment portfolios by balancing risk and return. In physics, they appear when analyzing circuits or mechanical systems where multiple forces or currents interact.
To solve the system of equations, we used substitution which led to approximate values of a ≈ 19.37832 and b ≈ 1.86486 . These values did not match any of the provided options. Hence, there is no correct choice among the options given.
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