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How do you solve the simultaneous equations $7 a - 3 b = 17$ $7a - 3b = 17$ and $2 a + b = 16$ $2a + b = 16$ ?

2 Answers

Jul 23, 2015

$(a, b) = (5, 6)$ $(a,b) = (5,6)$

Explanation:

[1] $XXXX$ $color(white)("XXXX")$ $7 a - 3 b = 17$ $7a-3b=17$
[2] $XXXX$ $color(white)("XXXX")$ $2 a + b = 16$ $2a+b =16$

Multiply both sides of [2] by $3$ $3$ (to get the same coefficient for $b$ $b$ as in [1])
[3] $XXXX$ $color(white)("XXXX")$ $6 a + 3 b = 48$ $6a+3b = 48$

Add [1] and [3]
[4] $XXXX$ $color(white)("XXXX")$ $13 a = 65$ $13a = 65$

Divide both sides by $13$ $13$
[5] $XXXX$ $color(white)("XXXX")$ $a = 5$ $a = 5$

Substitute $5$ $5$ for $a$ $a$ in [2]
[6] $XXXX$ $color(white)("XXXX")$ $2 (5) + b = 16$ $2(5) + b =16$

Simplify
[7] $XXXX$ $color(white)("XXXX")$ $b = 6$ $b = 6$

Jul 23, 2015

$a = 5$ $a=5$

$b = 6$ $b=6$

Explanation:

We are given $7 a - 3 b = 17$ $7a-3b=17$ and $2 a + b = 16$ $2a+b=16$

Let's get the $b$ $b$ 's equal to each other but opposite in sign by multiplying $2 a + b = 16$ $2a+b=16$ by $3$ $3$ :

$2 a + b = 16$ $2a+b=16$

$3 \cdot 2 a + 3 \cdot b = 3 \cdot 16$ $3*2a+3*b=3*16$

$6 a + 3 b = 48$ $6a+3b=48$

Now, let's add $6 a + 3 b = 48$ $6a+3b=48$ to $7 a - 3 b = 17$ $7a-3b=17$

$6 a + 3 b = 48$ $6a+3b=48$
$7 a - 3 b = 17$ $7a-3b=17$

$13 a = 65$ $13a=65$

$a = \frac{65}{13}$ $a=65/13$

$a = 5$ $a=5$

Now, substitute $a$ $a$ into $7 a - 3 b = 17$ $7a-3b=17$

$7 a - 3 b = 17$ $7a-3b=17$

$7 \cdot 5 - 3 b = 17$ $7*5-3b=17$

$35 - 3 b = 17$ $35-3b=17$

$- 3 b = 17 - 35$ $-3b=17-35$

$- 3 b = - 18$ $-3b=-18$

$b = 6$ $b=6$

Let's check to see if our answers are correct by plugging in the values we found for both $a$ $a$ and $b$ $b$ into $2 a + b = 16$ $2a+b=16$

$2 a + b = 16$ $2a+b=16$

$2 \cdot 5 + 6 = 16$ $2*5+6=16$

$10 + 6 = 16$ $10+6=16$

$16 = 16$ $16=16$

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