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How do you solve the following system using substitution?: $3 s - 5 t = - 30, 7 s + 11 t = 32$ $3s-5t=-30, 7s+11t=32$

1 Answer

Oct 23, 2015

$(s, t) = (- \frac{5}{2}, \frac{9}{2}) = (- 2 \frac{1}{2}, 4 \frac{1}{2})$ $(s,t)=(-5/2,9/2) = (-2 1/2, 4 1/2)$

Explanation:

Given
[1] $XXX 3 s - 5 t = - 30$ $color(white)("XXX")3s-5t=-30$
[2] $XXX 7 s + 11 t = 32$ $color(white)("XXX")7s+11t=32$

Rewriting [1] to generate an equation of the form: $s =$ $s =$ some expression
[3] $XXX 3 s = 5 t - 30$ $color(white)("XXX")3s = 5t-30$

[4] $XXX s = \frac{5 t - 30}{3}$ $color(white)("XXX")s = (5t-30)/3$

We can now substitute (as per the question) $(\frac{5 t - 30}{3})$ $((5t-30)/3)$ for $s$ $s$ in [2]
[5] $XXX 7 (\frac{5 t - 30}{3}) + 11 t = 32$ $color(white)("XXX")7((5t-30)/3)+11t = 32$

Simplifying
[6] $XXX \frac{35 t - 210 + 33 t}{3} = 32$ $color(white)("XXX")(35t-210 +33t)/3 = 32$

[7] $XXX (68 t - 210) = 96$ $color(white)("XXX")(68t-210) = 96$

[8] $XXX 68 t = 306$ $color(white)("XXX")68t = 306$

[9] $XXX t = \frac{9}{2}$ $color(white)("XXX")t = 9/2$

Substituting $\frac{9}{2}$ $9/2$ into [1] in place of $t$ $t$
[10] $XXX 3 s - 5 \cdot (\frac{9}{2}) = - 30$ $color(white)("XXX")3s -5*(9/2) = -30$

[11] $XXX 3 s = \frac{5 \cdot 9 - 2 \cdot 30}{2}$ $color(white)("XXX")3s = (5*9 -2*30)/2$

[12] $XXX 3 s = - \frac{15}{2}$ $color(white)("XXX")3s =-15/2$

[13] $XXX s = - \frac{5}{2}$ $color(white)("XXX")s = - 5/2$

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