Algebra Examples

$k = - 4 j - 7$ and $k = j^{2} - 3$

Step 1

Eliminate the equal sides of each equation and combine.

$- 4 j - 7 = j^{2} - 3$

Step 2

Solve $- 4 j - 7 = j^{2} - 3$ for $j$ .

Tap for more steps...

Step 2.1

Subtract $j^{2}$ from both sides of the equation.

$- 4 j - 7 - j^{2} = - 3$

Step 2.2

Add $3$ to both sides of the equation.

$- 4 j - 7 - j^{2} + 3 = 0$

Step 2.3

Add $- 7$ and $3$ .

$- 4 j - j^{2} - 4 = 0$

Step 2.4

Factor the left side of the equation.

Tap for more steps...

Step 2.4.1

Factor $- 1$ out of $- 4 j - j^{2} - 4$ .

Tap for more steps...

Step 2.4.1.1

Reorder $- 4 j$ and $- j^{2}$ .

$- j^{2} - 4 j - 4 = 0$

Step 2.4.1.2

Factor $- 1$ out of $- j^{2}$ .

$- (j^{2}) - 4 j - 4 = 0$

Step 2.4.1.3

Factor $- 1$ out of $- 4 j$ .

$- (j^{2}) - (4 j) - 4 = 0$

Step 2.4.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$- (j^{2}) - (4 j) - 1 \cdot 4 = 0$

Step 2.4.1.5

Factor $- 1$ out of $- (j^{2}) - (4 j)$ .

$- (j^{2} + 4 j) - 1 \cdot 4 = 0$

Step 2.4.1.6

Factor $- 1$ out of $- (j^{2} + 4 j) - 1 (4)$ .

$- (j^{2} + 4 j + 4) = 0$

Step 2.4.2

Factor using the perfect square rule.

Tap for more steps...

Step 2.4.2.1

Rewrite $4$ as $2^{2}$ .

$- (j^{2} + 4 j + 2^{2}) = 0$

Step 2.4.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 j = 2 \cdot j \cdot 2$

Step 2.4.2.3

Rewrite the polynomial.

$- (j^{2} + 2 \cdot j \cdot 2 + 2^{2}) = 0$

Step 2.4.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = j$ and $b = 2$ .

$- {(j + 2)}^{2} = 0$

Step 2.5

Divide each term in $- {(j + 2)}^{2} = 0$ by $- 1$ and simplify.

Tap for more steps...

Step 2.5.1

Divide each term in $- {(j + 2)}^{2} = 0$ by $- 1$ .

$\frac{- {(j + 2)}^{2}}{- 1} = \frac{0}{- 1}$

Step 2.5.2

Simplify the left side.

Tap for more steps...

Step 2.5.2.1

Dividing two negative values results in a positive value.

$\frac{{(j + 2)}^{2}}{1} = \frac{0}{- 1}$

Step 2.5.2.2

Divide ${(j + 2)}^{2}$ by $1$ .

${(j + 2)}^{2} = \frac{0}{- 1}$

Step 2.5.3

Simplify the right side.

Tap for more steps...

Step 2.5.3.1

Divide $0$ by $- 1$ .

${(j + 2)}^{2} = 0$

Step 2.6

Set the $j + 2$ equal to $0$ .

$j + 2 = 0$

Step 2.7

Subtract $2$ from both sides of the equation.

$j = - 2$

Step 3

Evaluate $k$ when $j = - 2$ .

Tap for more steps...

Step 3.1

Substitute $- 2$ for $j$ .

$k = {(- 2)}^{2} - 3$

Step 3.2

Simplify ${(- 2)}^{2} - 3$ .

Tap for more steps...

Step 3.2.1

Raise $- 2$ to the power of $2$ .

$k = 4 - 3$

Step 3.2.2

Subtract $3$ from $4$ .

$k = 1$

Step 4

The solution of the system of equations is all the values that make the system true.

$k = 1, j = - 2$