Finite Math Examples

$y = 4 x + 3 x - 2$ , $y = 5 x$

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Move all terms containing variables to the left side of the equation.

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Step 1.1.1

Subtract $4 x$ from both sides of the equation.

$y - 4 x = 3 x - 2$

$y = 5 x$

Step 1.1.2

Subtract $3 x$ from both sides of the equation.

$y - 4 x - 3 x = - 2$

$y = 5 x$

$y - 4 x - 3 x = - 2$

$y = 5 x$

Step 1.2

Subtract $3 x$ from $- 4 x$ .

$y - 7 x = - 2$

$y = 5 x$

Step 1.3

Reorder $y$ and $- 7 x$ .

$- 7 x + y = - 2$

$y = 5 x$

Step 1.4

Subtract $5 x$ from both sides of the equation.

$- 7 x + y = - 2$

$y - 5 x = 0$

Step 1.5

Reorder $y$ and $- 5 x$ .

$- 7 x + y = - 2$

$- 5 x + y = 0$

$- 7 x + y = - 2$

$- 5 x + y = 0$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array} |$

Step 3.2

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 7 \cdot 1 - (- 5 \cdot 1)$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $- 7$ by $1$ .

$- 7 - (- 5 \cdot 1)$

Step 3.3.1.2

Multiply $- (- 5 \cdot 1)$ .

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Step 3.3.1.2.1

Multiply $- 5$ by $1$ .

$- 7 - - 5$

Step 3.3.1.2.2

Multiply $- 1$ by $- 5$ .

$- 7 + 5$

Step 3.3.2

Add $- 7$ and $5$ .

$- 2$

$D = - 2$

Step 4

Since the determinant is not $0$ , the system can be solved using Cramer's Rule.

Step 5

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ .

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Step 5.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 2 & 1 \\ 0 & 1 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 2 \cdot 1 + 0 \cdot 1$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply $- 2$ by $1$ .

$- 2 + 0 \cdot 1$

Step 5.2.2.1.2

Multiply $0$ by $1$ .

$- 2 + 0$

Step 5.2.2.2

Add $- 2$ and $0$ .

$- 2$

$D_{x} = - 2$

Step 5.3

Use the formula to solve for $x$ .

$x = \frac{D_{x}}{D}$

Step 5.4

Substitute $- 2$ for $D$ and $- 2$ for $D_{x}$ in the formula.

$x = \frac{- 2}{- 2}$

Step 5.5

Divide $- 2$ by $- 2$ .

$x = 1$

Step 6

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ .

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Step 6.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 7 & - 2 \\ - 5 & 0 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 7 \cdot 0 - (- 5 \cdot - 2)$

Step 6.2.2

Simplify the determinant.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Multiply $- 7$ by $0$ .

$0 - (- 5 \cdot - 2)$

Step 6.2.2.1.2

Multiply $- (- 5 \cdot - 2)$ .

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Step 6.2.2.1.2.1

Multiply $- 5$ by $- 2$ .

$0 - 1 \cdot 10$

Step 6.2.2.1.2.2

Multiply $- 1$ by $10$ .

$0 - 10$

Step 6.2.2.2

Subtract $10$ from $0$ .

$- 10$

$D_{y} = - 10$

Step 6.3

Use the formula to solve for $y$ .

$y = \frac{D_{y}}{D}$

Step 6.4

Substitute $- 2$ for $D$ and $- 10$ for $D_{y}$ in the formula.

$y = \frac{- 10}{- 2}$

Step 6.5

Divide $- 10$ by $- 2$ .

$y = 5$

Step 7

List the solution to the system of equations.

$x = 1$

$y = 5$