Show and explain how replacing one equation by the
sum of that equation and a multiple of the other produces
a system with the same solutions as the one shown.
8x + 7y = 39
4x – 14y = -68
To solve the given system of equations, let's use the method of substitution:
Step 1: Solve one equation for one variable.
From the first equation, we can express y in terms of x:
Equation 1: -2x + y = 4
Adding 2x to both sides gives us:
y = 2x + 4
Step 2: Substitute the expression for y in the second equation.
Now, substitute the value of y from Equation 1 into the second equation:
x + 3(2x + 4) = 8
Simplifying the equation:
x + 6x + 12 = 8
7x + 12 = 8
Step 3: Solve for x.
To isolate x, subtract 12 from both sides:
7x = 8 - 12
7x = -4
Divide both sides by 7:
x = -4/7
Step 4: Substitute the value of x back into one of the original equations to solve for y.
Let's use Equation 1:
-2(-4/7) + y = 4
Simplifying the equation:
8/7 + y = 4
y = 4 - 8/7
y = 28/7 - 8/7
y = 20/7
So, the solution to the system of equations is:
x = -4/7
y = 20/7
Now, let's check our solution by substituting these values back into the original equations:
First equation: -2x + y = 4
-2(-4/7) + 20/7 = 4
8/7 + 20/7 = 4
28/7 = 4 (simplified)
4 = 4 (true)
Second equation: x + 3y = 8
-4/7 + 3(20/7) = 8
-4/7 + 60/7 = 8
56/7 = 8
8 = 8 (true)
Therefore, the solution x = -4/7, y = 20/7 satisfies both equations.
Now, let's determine the equivalent system of equations using the given theorem.
The theorem states that replacing one equation by the sum of that equation and a multiple of the other equation produces an equivalent system.
Let's rewrite the original system of equations:
Equation 1: -2x + y = 4
Equation 2: x + 3y = 8
To find an equivalent system, we can add Equation 1 to Equation 2 multiplied by a constant.
Let's consider the options one by one:
A) 7y = 20 (Equation 1) and x + 3y = 8 (Equation 2)
To create an equivalent system, we would need to multiply Equation 2 by (-2) and add it to Equation 1:
-2(x + 3y) + (7y) = -2(8) + 20
-2x - 6y + 7y = -16 + 20
-2x + y = 4
The resulting system is indeed equivalent to the original system.
Therefore, the correct answer is:
\[
\begin{array}{l}
\left\{\begin{array}{l}
7 y=20 \\
x+3 y=8
\end{array}\right.
\end{array}
\]
I hope this helps! If you have any further questions, please let me know.
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.....(1)
...(2)
...(3)
..(4)
in equation (2)
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in equation (1)
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