Math

QuestionFind the value of aa for the system of equations with infinite solutions: 12x+13y=16 \frac{1}{2} x+\frac{1}{3} y=\frac{1}{6} and ax+y=c a x+y=c .

Studdy Solution

STEP 1

Assumptions1. The system of equations is given as {1x+13y=16ax+y=c \left\{\begin{array}{c} \frac{1}{} x+\frac{1}{3} y=\frac{1}{6} \\ a x+y=c \end{array}\right. . The system has an infinite number of solutions.
3. aa and cc are constants.

STEP 2

For a system of linear equations to have an infinite number of solutions, the equations must be equivalent, meaning they represent the same line. So, we need to express the second equation in the form of the first one.

STEP 3

The first equation can be rewritten by multiplying all terms by6 to get rid of the fractions.
3x+2y=13x +2y =1

STEP 4

Now, we need to express the second equation in the same form. To do this, we can divide the second equation by cc.
acx+1cy=1\frac{a}{c}x + \frac{1}{c}y =1

STEP 5

For the two equations to be equivalent, the coefficients of xx and yy in both equations must be equal. So, we haveac=3\frac{a}{c} =31c=2\frac{1}{c} =2

STEP 6

We are only interested in the value of aa. So, we can solve the first equation from5 for aa.
a=3ca =3c

STEP 7

But we also know from the second equation in5 that c=12c = \frac{1}{2}. So, we can substitute this value into the equation from6 to find the value of aa.
a=3×12a =3 \times \frac{1}{2}

STEP 8

Calculate the value of aa.
a=3×12=32a =3 \times \frac{1}{2} = \frac{3}{2}So, the value of aa is 32\frac{3}{2}.

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