Solve 34 + 5y = 16; 9H + 3y = 14 | 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

$$34+5y=16; 9H+3y=14$$

Solve for y, H

$y = -\frac{18}{5} = -3\frac{3}{5} = -3.6$
$H = \frac{124}{45} = 2\frac{34}{45} \approx 2.755555556$

Solution Steps

Consider the first equation. Subtract $34$ from both sides.

$$5y=16-34$$

Subtract $34$ from $16$ to get $-18$.

$$5y=-18$$

Divide both sides by $5$.

$$y=-\frac{18}{5}$$

Consider the second equation. Insert the known values of variables into the equation.

$$9H+3\left(-\frac{18}{5}\right)=14$$

Multiply $3$ and $-\frac{18}{5}$ to get $-\frac{54}{5}$.

$$9H-\frac{54}{5}=14$$

Add $\frac{54}{5}$ to both sides.

$$9H=14+\frac{54}{5}$$

Add $14$ and $\frac{54}{5}$ to get $\frac{124}{5}$.

$$9H=\frac{124}{5}$$

Divide both sides by $9$.

$$H=\frac{\frac{124}{5}}{9}$$

Express $\frac{\frac{124}{5}}{9}$ as a single fraction.

$$H=\frac{124}{5\times 9}$$

Multiply $5$ and $9$ to get $45$.

$$H=\frac{124}{45}$$

The system is now solved.

$$y=-\frac{18}{5}$$ $$H=\frac{124}{45}$$