Question

If DF = 42, find DE

please help
:)​

Answer 1

we know the full length DF = 42, hmmm wait a second, but the full length is really the segments DE + EF, namely

`\bf \stackrel{DE}{(7x+1)}~~+~~\stackrel{EF}{(4x-3)}~~=~~\stackrel{DF}{42}\qquad \implies \qquad 11x-2=42 \\\\\\ 11x=44\implies x = \cfrac{44}{11}\implies x = 4 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{DE}{7x+1}\implies 7(4)+1\implies \stackrel{DE}{29}`

Answer 2

To find the value of DE when DF = 42, DE = 7x + 1, and EF = 4x - 3, solve the equation DE + EF = DF for x and then substitute back to get DE = 29.

If DF = 42, we want to find the value of DE where DE = 7x + 1 and EF = 4x - 3. We are given that DF is the sum of DE and EF, so the equation is:

DE + EF = DF

Substituting the given values into this equation, we get:

(7x + 1) + (4x - 3) = 42

Combining like terms results in:

11x - 2 = 42

Adding 2 to both sides of the equation:

11x = 44

Dividing both sides by 11:

x = 4

Now substituting x back into the expression for DE:

DE = 7(4) + 1

DE = 29

Therefore, DE is 29 units long based on the given linear equations.