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masha68 [24]
2 years ago
6

For what value of the constant a will the system of linear equations 6x-5y=3 and 3x+ay=1 have no solution

Mathematics
1 answer:
hodyreva [135]2 years ago
5 0

The value of the constant a will the system of linear equations 6x-5y=3 and 3x+ay=1 have no solution is -5/2

<h3>System of equation</h3>

For a system of equation to have no solution, the expression on both sides must be different.

Given the system of equation

6x-5y=3 and

3x+ay=1

For the equations to have no solution, the a1/a2= b1/b2

Substitute

6/3 = -5/a


Cross multiply

2 = -5/a

a= -5/2

Hence the value of the constant a will the system of linear equations 6x-5y=3 and 3x+ay=1 have no solution is -5/2

Learn more on system of equation here: brainly.com/question/14323743

#SPJ1

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Dafna1 [17]
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Exact Form:    ㏒ ( 24 )

Decimal Form:    1.38021124


<h2>Explanation:</h2><h3>                            Use the product property of logarithms,  </h3>

                                ㏒b ( x)  +  ㏒b ( y )  =  ㏒b ( x y ).

                                 ㏒ ( 6 ⋅ 8 ) −  ㏒ ( 2 ) .


<h3>                         ⇒Use the quotient property of logarithms,</h3>

                                 ㏒ b ( x ) − ㏒ b ( y ) = ㏒ b ( x y ) .

                                 ㏒ ( 6 ⋅ 8/ 2 )

<h3>                          ⇒Reduce the expression by cancelling the common factors. </h3>

                                Factor  2  out of  6 ⋅8 .

                                log ( 2 ( 3 ⋅ 8 )  / 2 )

<h3>                             Divide  3 ⋅ 8   by  1 .</h3>

                                ㏒ ( 3 ⋅ 8 )

                                Multiply  3  by  8 .

                                 ㏒ ( 24 )

<h3>The result can be shown in both exact and decimal forms. </h3><h3>                            Exact Form:  ㏒ ( 24 ) </h3><h3>                            Decimal Form:  1.38021124</h3>
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Step-by-step explanation:

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