Solve {l}{3(n-2)-19=5+2(n+5)}{3(4k-1)-(6k-10)=7k}

$$\left. \begin{array} { l } { 3 ( n - 2 ) - 19 = 5 + 2 ( n + 5 ) } \\ { 3 ( 4 k - 1 ) - ( 6 k - 10 ) = 7 k } \end{array} \right.$$

Solve for n, k

$$n=40$$

$$k=7$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 3 by n-2.

$$3n-6-19=5+2\left(n+5\right)$$

Subtract 19 from -6 to get -25.

$$3n-25=5+2\left(n+5\right)$$

Use the distributive property to multiply 2 by n+5.

$$3n-25=5+2n+10$$

Add 5 and 10 to get 15.

$$3n-25=15+2n$$

Subtract 2n from both sides.

$$3n-25-2n=15$$

Combine 3n and -2n to get n.

$$n-25=15$$

Add 25 to both sides.

$$n=15+25$$

Add 15 and 25 to get 40.

$$n=40$$

Consider the second equation. Use the distributive property to multiply 3 by 4k-1.

$$12k-3-\left(6k-10\right)=7k$$

To find the opposite of 6k-10, find the opposite of each term.

$$12k-3-6k+10=7k$$

Combine 12k and -6k to get 6k.

$$6k-3+10=7k$$

Add -3 and 10 to get 7.

$$6k+7=7k$$

Subtract 7k from both sides.

$$6k+7-7k=0$$

Combine 6k and -7k to get -k.

$$-k+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$-k=-7$$

Divide both sides by -1.

$$k=\frac{-7}{-1}$$

Fraction \frac{-7}{-1} can be simplified to 7 by removing the negative sign from both the numerator and the denominator.

$$k=7$$

The system is now solved.

$$n=40 k=7$$