Question Number 93057 by gonzo last updated on 10/May/20
$${y}={b}\mathrm{1}{x}\mathrm{1}+{b}\mathrm{2}{x}\mathrm{2}+{c} \\ $$$${i}\:{need}\:{to}\:{arrange}\:{the}\:{equation}\:{for}\:{thd}\:{value}\:{of}\:{x}\mathrm{2} \\ $$
Commented by gonzo last updated on 10/May/20
$${i}\:{have}\:{discalcula}\:{and}\:{it}\:{helps}\:{me}\:{work}\:{kt}\:{through} \\ $$
Answered by prakash jain last updated on 10/May/20
$${y}={b}_{\mathrm{1}} {x}_{\mathrm{1}} +{b}_{\mathrm{2}} {x}_{\mathrm{2}} +{c} \\ $$$$\mathrm{subtract}\:{b}_{\mathrm{1}} {x}_{\mathrm{1}} +{c}\:\mathrm{from}\:\mathrm{both}\:\mathrm{sides}. \\ $$$${y}−{b}_{\mathrm{1}} {x}_{\mathrm{1}} −{c}={b}_{\mathrm{1}} {x}_{\mathrm{1}} +{b}_{\mathrm{2}} {x}_{\mathrm{2}} +{c}−{b}_{\mathrm{1}} {x}_{\mathrm{1}} −{c} \\ $$$${y}−{b}_{\mathrm{1}} {x}_{\mathrm{1}} −{c}={b}_{\mathrm{2}} {x}_{\mathrm{2}} \\ $$$$\mathrm{divide}\:\mathrm{by}\:{b}_{\mathrm{2}} \\ $$$$\frac{{y}−{b}_{\mathrm{1}} {x}_{\mathrm{1}} −{c}}{{b}_{\mathrm{2}} }=\frac{{b}_{\mathrm{2}} {x}_{\mathrm{2}} }{{b}_{\left(\right.} } \\ $$$${x}_{\mathrm{2}} =\frac{{y}−{b}_{\mathrm{1}} {x}_{\mathrm{1}} −{c}}{{b}_{\mathrm{2}} } \\ $$
Commented by gonzo last updated on 10/May/20
$${thank}\:{you}\:{so}\:{much} \\ $$