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Algebra Examples

(x + 3) - y = 4

$(x + 3) - y = 4$ ,

x - 4 y = 10

$x - 4 y = 10$ ,

(- 2, - 3)

$(- 2, - 3)$

Step 1

Simplify

((- 2) + 3) - (- 3)

$((- 2) + 3) - (- 3)$ .

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Step 1.1

Multiply

- 1

$- 1$ by

- 3

$- 3$ .

(- 2) + 3 + 3 = 4

$(- 2) + 3 + 3 = 4$

(- 2) - 4 \cdot - 3 = 10

$(- 2) - 4 \cdot - 3 = 10$

Step 1.2

Add

- 2

$- 2$ and

3

$3$ .

1 + 3 = 4

$1 + 3 = 4$

(- 2) - 4 \cdot - 3 = 10

$(- 2) - 4 \cdot - 3 = 10$

Step 1.3

Add

1

$1$ and

3

$3$ .

4 = 4

$4 = 4$

(- 2) - 4 \cdot - 3 = 10

$(- 2) - 4 \cdot - 3 = 10$

4 = 4

$4 = 4$

(- 2) - 4 \cdot - 3 = 10

$(- 2) - 4 \cdot - 3 = 10$

Step 2

Since

4 = 4

$4 = 4$ , the equation will always be true.

Always true

(- 2) - 4 \cdot - 3 = 10

$(- 2) - 4 \cdot - 3 = 10$

Step 3

Simplify

(- 2) - 4 \cdot - 3

$(- 2) - 4 \cdot - 3$ .

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Step 3.1

Multiply

- 4

$- 4$ by

- 3

$- 3$ .

- 2 + 12 = 10

$- 2 + 12 = 10$

Always true

Step 3.2

Add

- 2

$- 2$ and

12

$12$ .

10 = 10

$10 = 10$

Always true

10 = 10

$10 = 10$

Always true

Step 4

Since

10 = 10

$10 = 10$ , the equation will always be true.

Always true

Always true

Step 5

The ordered pair is a solution to the system of equations.

(- 2, - 3)

$(- 2, - 3)$ is a solution

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⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$