Given that a, b, and c are non-zero real numbers and a + b ≠ 0, solve for x. ax + bx - c = 0

Mathematics

2 answers:

Stolb23 [73]3 years ago

4 0

$ax+bx-c=0\\
x(a+b)=c\\
x=\dfrac{c}{a+b}$

JulsSmile [24]3 years ago

3 0

$ax+bx-c=0 \ \ \ |+c \\
ax+bx=c \\
(a+b)x=c \ \ \ |\div (a+b), a+b \not= 0 \\
\boxed{x=\frac{c}{a+b}}$

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Find the zero of each function and state the multiplicity of each zero. Please show all steps.

vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

Step-by-step explanation:

1. y=(x+3)^3

$y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3$

Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1)

$y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.$

Zeros: x=2 multiplicity 2; x=1 multiplicity 1

3. y=(2x+3)(x-1)^2

$y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.$