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(2x+5)(-mx+9)=0
In the given equation,
m is a constant. If the equation has the solutions
x=-(5)/(2) and
x=(3)/(2), what is the value of
m ?

◻

$(2 x+5)(-m x+9)=0$ $\newline$ In the given equation, $m$ is a constant. If the equation has the solutions $x=-\frac{5}{2}$ and $x=\frac{3}{2}$ , what is the value of $m$ ? $\newline$ $\square$

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Full solution

Q.

(2 x+5)(-m x+9)=0

\newline

In the given equation,

m

is a constant. If the equation has the solutions

x=-\frac{5}{2}

and

x=\frac{3}{2}

, what is the value of

m

?

\newline

\square

Given Equation and Solutions: We are given the equation $(2x+5)(-mx+9)=0$ and the solutions $x=-\frac{5}{2}$ and $x=\frac{3}{2}$ . We will use these solutions to find the value of $m$ .
Substitute $x=-\frac{5}{2}$ : Since $(2x+5)(-mx+9)=0$ , we know that either $2x+5=0$ or $-mx+9=0$ for the solutions given.
Substitute $x=\frac{3}{2}$ : First, let's use the solution $x=-\frac{5}{2}$ and substitute it into $2x+5=0$ to see if it satisfies this part of the equation. $\newline$ $2\cdot\left(-\frac{5}{2}\right) + 5 = -5 + 5 = 0$ $\newline$ This shows that $x=-\frac{5}{2}$ is a solution for $2x+5=0$ .
Solve for m: Now, let's use the solution $x=\frac{3}{2}$ and substitute it into $-mx+9=0$ to find the value of m. $\newline$ $-m\cdot\frac{3}{2} + 9 = 0$ $\newline$ Multiplying both sides by $2$ to get rid of the fraction, we get: $\newline$ $-3m + 18 = 0$
Isolate $m$ : Now, we solve for $m$ by adding $3m$ to both sides of the equation: $\newline$ $-3m + 18 + 3m = 0 + 3m$ $\newline$ $18 = 3m$
Isolate m: Now, we solve for m by adding $3m$ to both sides of the equation: $\newline$ $-3m + 18 + 3m = 0 + 3m$ $\newline$ $18 = 3m$ Divide both sides by $3$ to isolate $m$ : $\newline$ $\frac{18}{3} = \frac{3m}{3}$ $\newline$ $m = 6$

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