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dimaraw [331]
3 years ago
8

2 Solve for k. -8k+ 6 = –10k + 10 k

Mathematics
2 answers:
Sever21 [200]3 years ago
8 0

Answer:

k=2

Step-by-step explanation:

tigry1 [53]3 years ago
4 0

Answer:

k=3/4

Step-by-step explanation:

-8k+6=-10k+10k

-8k+6=0

6=8k

8k=6

k=6/8

k=3/4

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I need help 6x-2y=10 x-2y=-5 solve by elimination
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<h3><u>Explanation</u></h3>
  • Given the system of equations.

\begin{cases} 6x - 2y = 10 \\ x - 2y =  - 5 \end{cases}

  • Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

\begin{cases} 6x - 2y = 10 \\  - x  +  2y =  5 \end{cases}

There as we can get rid of the y-term by adding both equations.

(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x =  \frac{15}{5}  \longrightarrow  \frac{ \cancel{15}}{ \cancel{5}}  =  \frac{3}{1}  \\ x = 3

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

x - 2y =  - 5 \\ 3 - 2y =  - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\  \frac{8}{2}  = y \\ y =  \frac{8}{2} \longrightarrow  \frac{ \cancel{8}}{ \cancel{2}}  =  \frac{4}{1}  \\ y = 4

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

  • Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10  = 10 \longrightarrow \sf{true} \:  \green{ \checkmark}

<u>Second</u><u> </u><u>Equation</u>

x - 2y =  - 5 \\ 3 - 2(4) =  - 5 \\ 3 - 8 =  - 5 \\  - 5 =  - 5 \longrightarrow  \sf{true} \:  \green{ \checkmark}

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

\begin{cases} x = 3 \\ y = 4 \end{cases} \\  \sf \underline{Coordinate \:  \: Form} \\ (3,4)

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