Which equation has a solution of k = 6.5?

1 answer:

5 0

K=6.5

We move all terms to the left:

k-(6.5)=0

We add all the numbers together, and all the variables

k-6.5=0

We move all terms containing k to the left, all other terms to the right

k=6.5

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Read 2 more answers

Solve each system using elimination. tell whether the system has one solution, infinitely many solutions, or no solution.

Tomtit [17]

1.
$x-2y=-1 \\
2x+y=4 \ \ \ |\times 2 \\ \\
x-2y=-1 \\
\underline{4x+2y=8 \ } \\
x+4x=8-1 \\
5x=7 \\
x=\frac{7}{5} \\
x=1 \frac{2}{5} \\ \\
x-2y=-1 \\
\frac{7}{5}-2y=-1 \\
-2y=-1-\frac{7}{5} \\
-2y=-\frac{5}{5}-\frac{7}{5} \\
-2y=-\frac{12}{5} \\
y=\frac{6}{5} \\
y=1 \frac{1}{5} \\ \\
\boxed{(x,y)=(1\frac{2}{5}, 1 \frac{1}{5}) \Leftarrow \hbox{one solution}}$

2.
$x+3y=4 \ \ \ |\times 2 \\
2x-6y=8 \\ \\
2x+6y=8 \\
\underline{2x-6y=8} \\
2x+2x=8+8 \\
4x=16 \\
x=4 \\ \\
x+3y=4 \\
4+3y=4 \\
3y=4-4 \\
3y=0 \\
y=0 \\ \\
\boxed{(x,y)=(4,0) \Leftarrow \hbox{one solution}}$

3.
$y=-\frac{1}{2}x-3 \ \ \ |\times (-2) \\
x+2y=-6 \\ \\
-2y=x+6 \\
\underline{x+2y=-6 \ } \\
x=x+6-6 \\
x-x=6-6 \\
0=0 \\ \\
\boxed{\hbox{infinitely many solutions}}$

4.
$6x-3y=-18 \\
-2x+4y=18 \ \ \ |\times 3 \\ \\
6x-3y=-18 \\
\underline{-6x+12y=54} \\
-3y+12y=54-18 \\
9y=36 \\
y=4 \\ \\
6x-3y=-18 \\
6x-3 \times 4=-18 \\
6x-12=-18 \\
6x=-18+12 \\
6x=-6 \\
x=-1 \\ \\
\boxed{(x,y)=(-1,4) \Leftarrow \hbox{one solution}}$

3 0

3 years ago

Once I get it wrong, I can’t redo the same question again and I have to go onto the next question. I only have 5 questions but i

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Answer:

$\Huge\boxed{29.3}$