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Math  /  Algebra

QuestionUse Cramer's rule to solve the system of equations. If D=0D=0, use another method to determine the solution set. 3x+4y=75x−7y=−2\begin{array}{l} 3 x+4 y=7 \\ 5 x-7 y=-2 \end{array}
Select the correct choice below and, if necessary, fill in the answer box to complete yoi choice. A. There is one solution. The solution set is \{ □\square \}. (Simplify your answer. Type an ordered pair.) B. The system has infinitely many solutions. The solution set is □\square \}. (Simplify your answer. Type an ordered pair.) C. There is no solution. The solution set is ∅\varnothing.

Studdy Solution

STEP 1

What is this asking? We've got two equations with two unknowns, xx and yy, and we need to find the values of xx and yy that satisfy both equations using Cramer's rule, or another method if Cramer's rule doesn't work. Watch out! Cramer's rule is awesome, but it has a little quirk.
If the determinant of the coefficient matrix is zero, it won't work!
So, we need to check that first.

STEP 2

1. Calculate the determinant D
2. Calculate xx
3. Calculate yy

STEP 3

Let's **build our coefficient matrix** from the coefficients of xx and yy in our equations.
It looks like this: [345−7]\begin{bmatrix} 3 & 4 \\ 5 & -7 \end{bmatrix}

STEP 4

Now, let's **calculate the determinant** DD.
Remember, it's the product of the main diagonal minus the product of the other diagonal. D=(3⋅−7)−(4⋅5)=−21−20=−41D = (3 \cdot -7) - (4 \cdot 5) = -21 - 20 = \mathbf{-41}

STEP 5

Since DD is **not zero**, Cramer's rule is good to go!

STEP 6

To find xx, we replace the xx coefficients in our coefficient matrix with the constants from the right-hand side of our equations.
This gives us the **x-matrix**: [74−2−7]\begin{bmatrix} 7 & 4 \\ -2 & -7 \end{bmatrix}

STEP 7

Now, we **calculate the determinant** DxD_x of this new matrix: Dx=(7⋅−7)−(4⋅−2)=−49+8=−41D_x = (7 \cdot -7) - (4 \cdot -2) = -49 + 8 = \mathbf{-41}

STEP 8

Finally, we **divide** DxD_x by DD to **find** xx: x=DxD=−41−41=1x = \frac{D_x}{D} = \frac{-41}{-41} = \mathbf{1}

STEP 9

Similar to finding xx, we replace the yy coefficients in our coefficient matrix with the constants to get the **y-matrix**: [375−2]\begin{bmatrix} 3 & 7 \\ 5 & -2 \end{bmatrix}

STEP 10

Now, we **calculate the determinant** DyD_y: Dy=(3⋅−2)−(7⋅5)=−6−35=−41D_y = (3 \cdot -2) - (7 \cdot 5) = -6 - 35 = \mathbf{-41}

STEP 11

Finally, we **divide** DyD_y by DD to **find** yy: y=DyD=−41−41=1y = \frac{D_y}{D} = \frac{-41}{-41} = \mathbf{1}

STEP 12

The solution to the system of equations is (1,1)(\mathbf{1}, \mathbf{1}).
So the answer is A.
There is one solution.
The solution set is {(1,1)}\{ (1, 1) \}.

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