if the point (3,4) lie on graph of linear equation 3y = kx +7 then find the values of k. Also find two more
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Answered by
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3×4 = k×3 + 7
12 = 3 k+ 7
12 - 7 = 3k
5 = 3k
5/3 = k or 1.6
On putting x = 0 then y is
3y = 1.6 × 0 + 7
3y = 0 + 7
y = 7/3
y = 2.3
On putting x = 1 then y is
3y = 1.6 ×1 +7
3y = 1.6 + 7
3y =8.6
y = 8.6/3
12 = 3 k+ 7
12 - 7 = 3k
5 = 3k
5/3 = k or 1.6
On putting x = 0 then y is
3y = 1.6 × 0 + 7
3y = 0 + 7
y = 7/3
y = 2.3
On putting x = 1 then y is
3y = 1.6 ×1 +7
3y = 1.6 + 7
3y =8.6
y = 8.6/3
dikky:
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Answered by
1
The value of k is 5/3.
Given,
A point (3,4) lies on the graph of the linear equation 3y = kx+7.
To Find,
The value of k.
Solution,
Since the point (3,4) lies on the graph of the linear equation 3y = kx+7, so on substituting the point in the equation it will satisfy the equation.
So,
Substituting the values
3(4) = k(3)+7
12 = 3k+7
3k = 5
k = 5/3
Hence, the value of k is 5/3.
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