How many solutions does this linear system have?
[tex]
\begin{array}{l}
y=\frac{2}{3} x+2 \\
6 x-4 y=-10
\end{array}
[/tex]
A. one solution: (-0.6,-1.6)
B. one solution: (-0.6,1.6)
C. no solution
D. infinite number of solutions
Answer (1)
Rewrite the first equation in standard form: − 2 x + 3 y = 6 .
The second equation is already in standard form: 6 x − 4 y = − 10 .
Compare the ratios of the coefficients: 6 − 2 = − 4 3 .
Since the ratios of the coefficients of x and y are not equal, the system has one solution: one solution: ( − 0.6 , 1.6 ) .
Explanation
Analyze the problem and data We are given the following system of linear equations:
Equation 1: y = 3 2 x + 2 Equation 2: 6 x − 4 y = − 10
We want to determine the number of solutions this system has. To do this, we will rewrite both equations in the standard form A x + B y = C and compare the coefficients.
Rewrite Equation 1 Rewrite Equation 1 in standard form:
Multiply both sides of y = 3 2 x + 2 by 3 to eliminate the fraction:
3 y = 2 x + 6
Rearrange the equation to get it in the form A x + B y = C :
− 2 x + 3 y = 6
Equation 2 in standard form Equation 2 is already in standard form:
6 x − 4 y = − 10
Compare the coefficients Now we compare the coefficients of the two equations:
Equation 1: − 2 x + 3 y = 6 Equation 2: 6 x − 4 y = − 10
We compute the ratios of the coefficients:
A 2 A 1 = 6 − 2 = − 3 1
B 2 B 1 = − 4 3 = − 4 3
C 2 C 1 = − 10 6 = − 5 3
Determine the number of solutions Since A 2 A 1 = B 2 B 1 , the lines intersect and there is exactly one solution.
Confirm the solution To confirm, we can solve the system. From Equation 1, y = 3 2 x + 2 . Substitute this into Equation 2:
6 x − 4 ( 3 2 x + 2 ) = − 10
6 x − 3 8 x − 8 = − 10
Multiply by 3 to eliminate the fraction:
18 x − 8 x − 24 = − 30
10 x = − 6
x = − 10 6 = − 5 3 = − 0.6
Now substitute x = − 0.6 into Equation 1:
y = 3 2 ( − 0.6 ) + 2
y = − 0.4 + 2 = 1.6
So the solution is ( − 0.6 , 1.6 ) .
Final Answer Therefore, the linear system has one solution: ( − 0.6 , 1.6 ) .
Examples
Understanding the number of solutions in a linear system is crucial in various real-world applications. For instance, when planning a budget, you might have two equations representing your income and expenses. If the system has one solution, it means your budget is balanced at a specific point. If there are no solutions, your expenses exceed your income, indicating a need for adjustments. If there are infinite solutions, it suggests flexibility in your budget, allowing for various spending and saving scenarios. This concept extends to resource allocation, engineering designs, and economic modeling, where understanding the solution space helps in making informed decisions.
