Question
$(b+c)(y+z)-a x=b-c$,$(c+a)(z+x)-b y=c-a$$(a+b)(x+y)-c z=a-b$,where $a+b+c \neq 0$, then $x=$(A) $\frac{c-b}{a+b+c}$(B) $\frac{a-c}{a+b+c}$(C) $\frac{b-a}{a+b+c}$(D) $\frac{1}{a+b+c}$
Step 1
\end{align*} Show more…
Show all steps
Your feedback will help us improve your experience
Ahmad Reda and 67 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$(b+c)(y+z)-a x=b-c$, $(c+a)(z+x)-b y=c-a$, $(a+b)(x+y)-c z=a-b$, where $a+b+c \neq 0$, then $x=$ (A) $\frac{c-b}{a+b+c}$ (B) $\frac{a-c}{a+b+c}$ (C) $\frac{b-a}{a+b+c}$ (D) $\frac{1}{a+b+c}$
Given the system of equations $$ \begin{aligned} &(b+c)(y+z)-a x=b-c \\ &(c+a)(z+x)-b y=c-a \\ &(a+b)(x+y)-c z=a-b \end{aligned} $$ (where $a+b+c \neq 0) ;$ then $x: y: z$ is given by (a) $b-c: c-a: a-b$ (b) $b+c: c+a: a+b$ (c) $a: b: c$ (d) $\frac{a}{b}: \frac{b}{e^{2}}: \frac{c}{a}$
Challenge Problem Solve for $x, y,$ and $z,$ assuming $a \neq 0, b \neq 0,$ and $c \neq 0$ $$ \left\{\begin{array}{l} a x+b y+c z =a+b+c \\ a^{2} x+b^{2} y+c^{2} z =a c+a b+b c \\ a b x+b c y \quad \quad=b c+a c \end{array}\right. $$
Systems of Equations and Inequalities
Systems of Linear Equations: Substitution and Elimination
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD