Solve 10x + 5y + 5z = 50000; 15x + 4y + 7z = 63000; 13x + 3y + 8z = 59000 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$10x+5y+5z=50000; 15x+4y+7z=63000; 13x+3y+8z=59000$$

Solve for x, y, z

$x=2000$
$y=3000$
$z=3000$

Short Steps Using Substitution

Solve $10x+5y+5z=50000$ for $x$.

$$x=5000-\frac{1}{2}y-\frac{1}{2}z$$

Substitute $5000-\frac{1}{2}y-\frac{1}{2}z$ for $x$ in the second and third equation.

$$15\left(5000-\frac{1}{2}y-\frac{1}{2}z\right)+4y+7z=63000$$ $$13\left(5000-\frac{1}{2}y-\frac{1}{2}z\right)+3y+8z=59000$$

Solve these equations for $y$ and $z$ respectively.

$$y=\frac{24000}{7}-\frac{1}{7}z$$ $$z=-4000+\frac{7}{3}y$$

Substitute $\frac{24000}{7}-\frac{1}{7}z$ for $y$ in the equation $z=-4000+\frac{7}{3}y$.

$$z=-4000+\frac{7}{3}\left(\frac{24000}{7}-\frac{1}{7}z\right)$$

Solve $z=-4000+\frac{7}{3}\left(\frac{24000}{7}-\frac{1}{7}z\right)$ for $z$.

$$z=3000$$

Substitute $3000$ for $z$ in the equation $y=\frac{24000}{7}-\frac{1}{7}z$.

$$y=\frac{24000}{7}-\frac{1}{7}\times 3000$$

Calculate $y$ from $y=\frac{24000}{7}-\frac{1}{7}\times 3000$.

$$y=3000$$

Substitute $3000$ for $y$ and $3000$ for $z$ in the equation $x=5000-\frac{1}{2}y-\frac{1}{2}z$.

$$x=5000-\frac{1}{2}\times 3000-\frac{1}{2}\times 3000$$

Calculate $x$ from $x=5000-\frac{1}{2}\times 3000-\frac{1}{2}\times 3000$.

$$x=2000$$

The system is now solved.

$$x=2000$$ $$y=3000$$ $$z=3000$$