Solve 34 + y - 2 = 0; 2H + 3y - 4 = 0 | 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

$$34+y-2=0; 2H+3y-4=0$$

Solve for y, H

$y=-32$
$H=50$

Solution Steps

Consider the first equation. Subtract $2$ from $34$ to get $32$.

$$32+y=0$$

Subtract $32$ from both sides. Anything subtracted from zero gives its negation.

$$y=-32$$

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

$$2H+3\left(-32\right)-4=0$$

Multiply $3$ and $-32$ to get $-96$.

$$2H-96-4=0$$

Subtract $4$ from $-96$ to get $-100$.

$$2H-100=0$$

Add $100$ to both sides. Anything plus zero gives itself.

$$2H=100$$

Divide both sides by $2$.

$$H=\frac{100}{2}$$

Divide $100$ by $2$ to get $50$.

$$H=50$$

The system is now solved.

$$y=-32$$ $$H=50$$