Solve {l}{D+T+A=134}{T=(4)A}{A=8-D}

$$\left. \begin{array} { l } { D + T + A = 134 } \\ { T = ( 4 ) A } \\ { A = 8 - D } \end{array} \right.$$

Solve for D, T, A

$$D = -\frac{47}{2} = -23\frac{1}{2} = -23.5$$

$$T=126$$

$$A = \frac{63}{2} = 31\frac{1}{2} = 31.5$$

Short Steps Using Substitution

Reorder the equations.

$$T=4A D+T+A=134 A=8-D$$

Substitute 4A for T in the equation D+T+A=134.

$$D+4A+A=134$$

Solve D+4A+A=134 for D.

$$D=-5A+134$$

Substitute -5A+134 for D in the equation A=8-D.

$$A=8-\left(-5A+134\right)$$

Solve A=8-\left(-5A+134\right) for A.

$$A=\frac{63}{2}$$

Substitute \frac{63}{2} for A in the equation D=-5A+134.

$$D=-5\times \frac{63}{2}+134$$

Calculate D from D=-5\times \frac{63}{2}+134.

$$D=-\frac{47}{2}$$

Substitute \frac{63}{2} for A in the equation T=4A.

$$T=4\times \frac{63}{2}$$

Calculate T from T=4\times \frac{63}{2}.

$$T=126$$

The system is now solved.

$$D=-\frac{47}{2} T=126 A=\frac{63}{2}$$